Polymers Revision Video – Class 12, JEE, NEET

Polymers are defined as high molecular mass macromolecules which consist of repeating structural units derived from the appropriate monomers. The unit molecules that combine with each other to to form a polymer is called a monomer.

Lets have a quick glance of Class 12th Chapter Polymers with Prateek Gupta Sir (IIT BOMBAY) in one shot video:

CLICK HERE FOR COMPLETE PHYSICS REVISION | Class 11, 12, JEE, NEET

Mind Maps for Hydrocarbon – Alkyne Revision | Class 12, JEE, NEET

Here are the mind maps to help you in Class 12 Hydrocarbon Alkyne Revision. Important Reaction Mechanisms,  key points, etc in the mind map to help in quick revision.

#### MIND MAP 2

Mind Map for Biomolecules Part 2 Revision – Class 12, JEE, NEET

Here are the mind maps for to revise important reaction mechanisms, key concepts and more in Biomolecules.

#### MIND MAP – 3

Mind Map for Biomolecules Part 1 Revision – Class 12, JEE, NEET

Here are the mind maps for to revise important reaction mechanisms, key concepts and more in Biomolecules.

#### MIND MAP – 4

Organic Chemistry Revision Series for JEE, NEET, Class 11 and 12

Revise complete organic chemistry on eSaral YouTube Channel. Get everything you are looking for JEE and NEET Preparation in this Organic Chemistry Revision Series by Prateek Gupta Sir (IIT-BOMBAY). Here you will get to know about important reaction mechanisms and all the important concepts that are very important to solve conceptual problems that will come in main Exam.

These revision Videos also covers some tricks and tips to solve Multiple Choice questions of JEE & NEET in less time.

### Class XI & XII

TOPIC Revision Videos Mind Maps
General Organic Chemistry Revise Mindmap
Acidic and Basic Strength Revise Mindmap
Hydrocarbons Revise Mindmap
Optical Isomerism Revise Mindmap
Haloalkane & Haloarenes Revise Mindmap
Alcohol, Phenol and Ethers Revise Mindmap
Carbonyl Compounds Revise Mindmap
Aldol, Cannizzaro and Haloform Reaction Revise Mindmap
Name Reaction Revise Mindmap
Oxidation Reaction Revise Mindmap
Reduction Reaction Revise Mindmap
Carboxylic Acid Revise Mindmap
Nitrogen Containing Compounds Revise Mindmap
Bio-molecules Revise Mindmap
Polymers Revise Mindmap
Chemistry in Everyday Life Revise Mindmap

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Mindmaps for Hydrocarbons Revision – Class 12, JEE , NEET

CBSE class 11 hydrocarbons teach about alkanes, preparation of alkanes from unsaturated hydrocarbons, from alkyl halides, from carboxylic acids, physical and chemical properties of alkanes. Hydrocarbons topic gives knowledge about Alkenes, the nomenclature of alkenes, isomers, preparation of alkene using alkyne.

Here are the mind maps for to revise important reaction mechanisms, key concepts and more in hydrocarbons.

### Aromatic Hydrocarbons

CLICK HERE for Complete Physics Revision | Class 11, 12, JEE, NEET

Alcohol, Phenol and Ether Revision Video – Class 12, JEE, NEET

Alcohol, Phenol, and Ether are classes of organic compounds. These compounds have huge applications in industries for domestic purposes. When hydroxyl (-OH) group bonds with saturated carbon atom we get Alcohol. And dehydration of alcohol forms Ether. Monohydric, Dihydric, and Trihydric are three types of alcohols, based on the hydroxyl group.

Revise complete Alcohol, Phenol and Ethers with Prateek Gupta Sir (IIT-Bombay). The Revision of this chapter is divided into 3 parts:

### PART 3 – Phenols

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Haloalkane and Haloarenes Revision Video – Class 12, JEE, NEET

Haloalkanes and haloarenes are the hydrocarbons in which one or more hydrogen atoms have been replaced with halogen atoms. The primary difference between haloalkanes and haloarenes is that haloalkanes are derived from open-chain hydrocarbons (alkanes) whereas haloarenes are derived from aromatic Hydrocarbons.

### Haloarenes

CLICK HERE for Complete Physics Revision | Class 11, 12, JEE, NEET

ATOMIC THEORY OF MAGNETISM || Magnetism and Matter Class 12, JEE & NEET

Magnetism and Matter Class 12 Notes will help you in your Exam Preparation and will also help in scoring good! The property of any object by virtue of which it can attract a piece of iron or steel is called magnetism. Here will study about the Atomic Theory of Magnetism.

Atomic Theory of Magnetism :

(1) Each atom behaves like a complete magnet having a north and south pole of equal strength. The electrons revolving around the nucleus in an atom are equivalent to small current loops which behaves as magnetic dipole.

(2) In unmagnetized magnetic substance these atomic magnets (represented by arrows) are randomly oriented and form closed chains. The atomic magnets cancel the effect of each other and thus resultant magnetism is zero.

1. In magnetised substance all the atomic magnets are aligned in same direction and thus resultant magnetism is non-zero.

The atomic theory explains the following facts in magnetism.

(1) Non existence of monopoles. The magnetic poles always exist in pairs and are of equal strength.

(2) When a magnet breaks than each part behaves like a complete magnet.

(3) Magnetisation of an electromagnet can be explained as alignment of atomic magnets in direction of magnetic field.

(4) This explains the phenomenon of saturation magnetization i.e. acquired magnetism remains constant even on increasing the external magnetizing field.

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Mind Maps for Photo Electric Effect Revision – Class XII, JEE, NEET

Photoelectric Effect in Class 12 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

Important Questions of PCMB Boards

Get important questions for Boards exams. Download or View the Important Question bank for all Subjects. These important questions will play significant role in clearing concepts. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 & Class 12.

# Class 11 PHYSICS

Rotational Motion

Gravitation

Simple Harmonic Motion

Circular Motion

Friction

Newton’s Laws Of Motion

Conservation Of Linear Of Momentum

Projectile Motion

Unit And Dimension

Heat And Thermodynamics

General Properties Of Matter

Motion In a Straight Line

# Class 12 PHYSICS

Electrostatics

Moving Charges & Magnetism

Wave Optics

Ray Optics

Semiconductor

Modern Physics

Alternating Current

Electromagnetic Induction

Capacitor

Current Electricity

# Class 11 CHEMISTRY

General Organic Chemistry

s-block

Chemical Bonding

Mole Concept

Boron Family

Carbon Family

Ionic Equilibrium

Electrochemistry

Periodic Properties

Thermodynamics and Thermochemistry

Chemical Equilibrium

Gaseous State

Atomic Structure

# Class 12 CHEMISTRY

Solution & Colligative Properties

Solid State

Chemical Kinetics

Carboxylic acids & Derivative

Aliphatic Hydrocarbon

Surface Chemistry

Co-ordination Compounds

Metallurgy

d & f block Elements

Noble Gases

You can access free study material for all three subject’s Physics, Chemistry and Mathematics.

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Solid State Short Notes for Class 12, IIT-JEE & NEET

eSaral provides chemistry short notes for JEE and NEET to help students in revising topics quickly. These notes are completely based on latest syllabus and it includes all the tips and tricks that will help you in learning chemistry better and score well.

The Notes will help you to understand the important topics and remember the key points for exam point of view.

You can also access detailed Notes of chemistry here.

Solution & Colligative Properties | Question Bank for Class 12 Chemistry

Get Solution & Colligative Properties important questions for Boards exams. Download or View the Important Question bank for Class 12 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 12 chemistry chapter wise CBSE.

Click Here for Detailed Chapter-wise Notes of Chemistry for Class 12th, JEE & NEET.

You can access free study material for all three subject’s Physics, Chemistry and Mathematics.

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Q. Why does molality of a solution remain unchanged with the rise in temperature ?

[AI 2004 C]

Sol. Mass of solvent does not change with temperature.

Q. What is van’t Hoff equation for dilute solution ?

Sol. $\pi V=n R T$ where $n$ is the no. of moles of solute present in $V$ litres of solution, $\pi$ is osmotic pressure, $T$ is temperature and $R$ is gas constant or solution constant.

Q. Mention a large scale use of the phenomenon called ‘reverse osmosis’.

[Delhi 2004; CBSE 2004]

Sol. It is used for desalination of sea water.

Q. What do you understand by ‘colligative properties’?

[Delhi 1999]

Sol. Colligative properties are those properties which depends upon number of particles of solute and not on natures of solute.

Q. Why is the cooking temperature in pressure cooker higher than in open pan ?

[Delhi 2002 C]

Sol. In pressure cooker, high pressure is exerted by steam due to which boiling point becomes higher than in open pan.

Q. What is it that elevation in boiling point of water is not same in following solutions? 0.1 molar $N a C l$ solution and 0.1 molar sugar solution.

Sol. $0.1 M N a C /$ has higher elevation in boiling point because it contains more number of particles than $0.1 M$ solution of glucose. $N a C l \rightarrow N a^{+}+C l^{-}$

Q. What is de-icing agent ? How does it work ?

Sol. Common salt is called de-icing agent because it lowers the freezing point of water to such an extent that it does not freeze to form ice. Hence, it is used to clear snow from roads.

Q. Why is the vapour pressure of a solution of glucose in water lower than that of water ?

[A.I.S.B. 2006]

Sol. At a number of sites, non-volatile glucose molecules will occupy the surface instead of water molecules. Hence, effective surface area for evaporation decreases and so the vapour pressure is lower.

Q. Give one example each of solid in gas and liquid in gas solutions.

[H.P.S.B. 1998]

Sol. Iodine vapour in air, aerated drinks.

Q. What happens when blood cells are placed in pure water ?

[Delhi 2003, 04; D.S.B. 2004 C]

Sol. Due to osmosis, water molecules move into blood cells through the cells walls. As a result, blood cells swell and may even burst.

Q. Write expression for the Raoult’s law for non-volatile solutes.

Sol. $\left(p^{\circ}-p_{s}\right) / p^{\circ}=n_{2} /\left(n_{1}+n_{2}\right)$ where $p^{\circ}=V . P .$ of pure solvent $p_{s}=V . P .$ of solution, $n_{2}=$ no. of moles of solute and $n_{1}=$ no. of moles of solvent.

Q. What is the difference between lowering of vapour pressure and relative lowering of vapour pressure ?

Sol. Lowering of V.P. $=p^{\circ}-p_{s} .$ Relative lowering of V.P. $=\left(p^{\circ}-p_{s}\right) / p^{\circ}$

Q. State any two characteristic of ideal solution.

[Delhi 1999; Foreign 1999]

Sol. Ideal solutions

(ii) They can be separated by fractional distillation

Q. On mixing equal volumes of water and ethanol, what type of deviation would you expect from Raoult’s law ?

[Delhi 1999 C]

Sol. It is because of $H$ -bonding between acetone and $C H C l_{3}$, the force of attraction increases, therefore energy is released.

Q. Why are ethers not miscible in water ?

Sol. Because they can not form H-bonds with water.

Q. Why is benzene insoluble in water but soluble in toluene?

[AI 1999 C]

Sol. Benzene is insoluble in water because benzene is non polar where as water is polar. Toluene is non polar solvent therefore, Benzene is soluble in toluene.

Q. Give an example of a compound in which hydrogen bonding results in the formation of a dimer.

Sol. Acetic acid forms dimer due to H-bonding.

Q. Why is freezing point depression 0.1 M sodium chloride solution nearly twice that of 0.1 M glucose solution ?

[A.I.S.B. 2006]

Sol. $\mathrm{NaCl},$ being an electrolytes, dissociates almost completely to give $N a^{+}$ and $\quad C I^{-}$ ions whereas glucose, being nonelectrolyte, does not dissociate. Hence, the number of particles in 0.1 MNaCl solution is nearly double than in

$0.1 M$ glucose solution. Freezing point depression, being a colligative properly, is therefore, nearly twice for solution than for glucose solution of same molarity.

Q. State the formula relating pressure of a gas with its mole fraction in a liquid solution in contact with it.

[DSB2005]

Sol. According to Henry’s law.

Partial pressure of gas above the solution

$=k_{H} \times$ mole fraction of the gas in the solution

or $\quad p_{A}=k_{H} \times x_{A^{\prime}}$ where $k_{H}=$ Henry’s constant.

Q. When fruits and vegetables are dried and placed in water, they slowly swell and return to original form. Why ? Does an increase in temperature accelerate the process ? Explain.

Sol. When fruits and vegetables that have dried and are placed in water osmosis takes place, i.e., water molecules pass through semipermiable membranes present in cell walls, therefore, they swell. If temperature is increased, osmosis will be faster.

Q. Why does the use of pressure cooker reduce cooking time?

Sol. At higher pressure over the liquid (due to weight of the pressure cooker lid), the liquid boils at higher temperature. Therefore, cooking occurs faster.

Q. Sodium chloride solution freezes at lower temperature than water but boils at higher temperature than water. Explain.

Sol. Freezing point of a liquid depresses on the addition of a non-volatile solute and therefore, a solution of sodium chloride freezes at lower temperature than freezing point of water. On he other hand, there is elevation in boiling point on the addition of a non-volatile solute and consequently boiling point of sodium chloride solution is more than that of water.

Q. Addition of $\mathrm{HgI}_{2}$ to aqueous solution of $K I$ shows an increase in vapour pressure. Why?

Sol. HgI$_{2}$ forms a complex with $K I$ and therefore, the number of particles in solution increases.

$H g I_{2}+2 K I \rightarrow K_{2}\left[H g I_{4}\right] \Longrightarrow 2 K^{+}+\left[H g I_{4}\right]^{2-}$

Therefore, the osmotic pressure increases.

Q. Give one example each of miscible liquid pairs showing positive and negative deviations from Raoult’s law. Give one reason each for such deviations.

[Delhi 2000; AI 2000]

Sol. Ethanol and water, methanol and $H_{2} O$ show positive deviation from Raoult’s law because force of attraction between them decreases on mixing. and and, show negative deviation from Raoult’s law because force of attraction between them increases on mixing.

Q. “The solution of a non-volatile solute boils at a higher temperature than the pure solvent.” Show this relationship on a graphic diagram.

[AI 2002 C]

Sol. The diagram shows that solution containing non-volatile solute boils at temperature $T_{2}$ which is higher than boiling point of pure solvent, i.e., $T_{1}$

Variation of vapour pressure with Temperature showing boiling point of solution is higher than that of pure solvent.

Q. Carbon tetrachloride and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structures of these compounds.

[AI 2003]

Sol. Carbon tetrachloride is a non-polar compounds, it cannot form $H$ -bond with water, that is why $C C l_{4}$ and water do not mix with each other. Ethanol is a polar compound and $H_{2} \mathrm{O}$ is polar solvent, there is $H$ -bond between ethanol and water, therefore, ethanol and water are miscible in all proportions.

Q. Calculate molarity and molality of a $13 \%$ solution (by weight) of sulphuric acid? Its density is $1.020 \mathrm{g} \mathrm{cm}^{-3}$

(Atomic mass $H=1, O=16, S=32 \text { a. } m . u .)$

[Foreign 1999]

Sol. $M=\frac{w_{B}}{M_{B}} \times \frac{1000}{w_{A}}=\frac{13 \times 1000}{98 \times \frac{100}{1.02}}=\frac{13 \times 10 \times 1.02}{98}=\frac{1326}{98}$

$=1.353 \mathrm{mol} L^{-1}$ or $1.353 \mathrm{M}$

$m=\frac{w_{B}}{M_{B}} \times \frac{100}{w_{A}}=\frac{13}{98} \times \frac{1000}{87}=1.524 \mathrm{mol} / \mathrm{kg}_{\mathrm{or}} 1.524 \mathrm{m}$

Q. A solution is prepared by adding 60 g of methyl alcohol to 120 g of water. Calculate the mole fraction of methanol and water.

[H.P. 1995]

Sol. Mass of methanol = 60 g

Moles of methanol $\left.=\frac{60}{32}=1.875 \text { (Molar mass }=32\right)$

Moles of water $=\frac{120}{18}=6.667$

Total No. of moles = 1.875 + 6.667 = 8.542

Mole fraction of methanol $=\frac{1.875}{8.542}=0.220$

Mole fraction of water $=\frac{6.667}{8.542}=0.780$

Q. Vapour pressure of pure water at $35^{\circ} \mathrm{C}$ is $31.82 \mathrm{mm} \mathrm{Hg}$. When $27.0 \mathrm{g}$ of solute is dissolved in $100 \mathrm{g}$ of water (at the same temperature) vapour pressure of the solution thus formed is $30.95 \mathrm{mm}$ Hg. Calculate the molecular mass of the solute.

Sol. $\frac{P_{A}^{\circ}-P_{A}}{P_{A}^{\circ}}=X_{B} \Rightarrow \frac{31.82-30.95}{31.82}=\frac{27 / M}{100 / 18}$

$\Rightarrow \frac{0.87}{31.82}=\frac{27}{M_{B}} \times \frac{18}{100} \Rightarrow M_{B}=\frac{27 \times 18 \times 31.82}{0.87 \times 100}$

$\Rightarrow 177.75 g \mathrm{mol}^{-1}$

Q. In a solution of urea, $3.0 g$ of it is dissolved in $100 \mathrm{ml}$ of water. What will be the freezing point of this solution? State the approximation made if any.

$\left[k_{f} \text { for water }=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}, \text { molar mass of urea }=60 \mathrm{g}\right.$ $\left.m o l^{-1}\right]$

[Delhi 2002 C]

Sol. $\Delta T_{f}=k_{f} \times m$

$\Delta T_{f}=1.86 \times \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}=\frac{1.86 \times 3}{60} \times \frac{1000}{100}=0.93$

Freezing point of solution $=273 K-0.93=272.07 K$

Q. Give one example each of miscible liquid pairs showing positive and negative deviation from Raoults law. give one reason each for such deviations.

[CBSE 2004]

Sol. Solution of $n$ hexane and ethanol shows positive deviations from Raoults law because $n$ -hexane molecular weaken the H-bonds between ethanol molecules which increases its vapour

Solution of acetone and chloroform shows negative deviations from Raoult’s law because of the formation of the $H$ -bonds between acetone and chloroform molecules.

Q. What is meant by abnormal molecular mass of solute ? Discuss the factors, which bring abnormality in the experimentally determined molecular masses of solutes using colligative properties.

[AI 1999 C]

Sol. The molecular mass obtained with the help of colligative property sometimes is different from normal molecular mass, it is called abnormal molecular mass. The factor which bring abnormality are

(i) Association : When solute particles undergo association, number of particles become less and molecular mass determined with the help of colligative property will be more.

(ii) Dissociation : It leads to increase in number of particles, therefore, increase in colligative property, therefore, therefore, decrease in molecular weight because colligative property is inversely proportional to molecular weight.

Q. Explain why vapour pressure of solvent is lowered by addition of a non-volative solute ?

Sol. It is observed that the presence of a non-volatile solute in a solution reduces the escaping tendency of solvent molecules into vapour phase. It is because some of the solute particles occupy the position of the solvent molecules on the liquid surface and thus dower the vapour pressure of the solvent.

Q. Ristinguish between ideal and non ideal solutions.

[MP 2001, 02, 06]

Sol.

Q. How will you determine the molecular mass of a substance by osmotic pressure method ?

Sol. According to Van’t Hoff equation

$\pi v=n R T$ …..(i)

$\pi=$ osmotic pressure; $T=$ Temperature for a solution

If $w$ gram of solute is dissolved in $v$ litres of he solution and $M$ is the molecular mass of the solute, then

$n=\frac{w}{M}$

Substituting this value in equation (v) we get,

$\pi v=\frac{w}{M} R T \Rightarrow M=\frac{w R T}{\pi v}$ …..(ii)

Thus, measuring the osmotic pressure of a solution containing grams of the solute in litres of the solution, at temperature the molecular man, of the solute can be calculated using equation (ii).

Q. The vapour pressure of pure liquid $A$ and $B$ are $70 \mathrm{mm}$ and $90 \mathrm{mm}$ Hg respectively at $25^{\circ} .$ The mole fraction of ^{ } $A^{\prime}$ in a solution of the two is $0.3 .$ Assuming that $A$ and $B$ form an ideal solution, calculate the partial pressure of each component in equilibrium with the solution.

[AI 1999 C]

Sol.

Q. Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving $6.00 \mathrm{g}$ Glaubers salt, $N a_{2} S O_{4}, 10 H_{2} O$ in. 100 kg of $H_{2} O\left[K_{f}=1.86\right.$ $\left.K \mathrm{kg} \mathrm{mol}^{-1}\right)$

[CBSE Board 1998]

Sol.

Q. $2 \mathrm{g}$ of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$ dissolved in $25 \mathrm{g}$ of benzene shows a depression in freezing point equal to $1.62 \mathrm{K} .$ Molar depression constant for benzene is $4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} .$ What is the percentage association of acid if it exists as dimer in solution.

[CBSE 2004]

Sol. Here $W_{A}=25 g, W_{b}=2 g, \Delta T_{f}=1.62 K, K_{f}=4.9 K \mathrm{kg} \mathrm{mol}^{-1}$

Observed molecular mass of benzoic acid,

$M_{B}=\frac{1000 K_{f} \times W_{B}}{\Delta T_{f} \times W_{A}}=\frac{1000 \times 4.9 \times 2}{1.62 \times 25}=242$

Calculated molecular mass of benzoic acid

$=72+5+12+32+1=122$

Van’t Hoff factor

$i=\frac{\text { Calculated molecular mass }}{\text { Observed molecular mass }}=\frac{122}{242}=.504$

Let $\alpha$ be the degree of dissociation

$2 C_{6} H_{5} C O O H \rightleftharpoons\left(C_{6} H_{5} C O O H\right)_{2}$

Total number of moles after association

$=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$

$\therefore \quad i=\frac{1-\frac{\alpha}{2}}{1}=.504$

or $\quad \alpha=(1-.504) \times 2=.496 \times 2=.992$

percentage of association $=99.2 \%$

Q. With the help of a simple diagram, describe Berkeley and Hartley’s method of determining the osmotic pressure of a dilute solution.

[MP 2002, 04, 06]

Sol. Experimental Measurement of Osmotic Pressure. A number of methods are available for measurement of osmotic pressure. The best out of these is Berkeley and Hartley’s Method.

This method is based upon applying pressure on the solution which is just sufficient to prevent the entry of the solvent into the solution through the semi-permeable membrane. The apparatus used by Berkeley and nartley is shown in Figure.

It consists of a porous tube (open at both ends) containing the semi-permeable membrance of copper ferrocyanide. The porous tube is fitted with a reservoir R on one side and a tube T on the other.

The porous tube is filled with the pure solvent so that the level in the tube T stands at the mark M. The porous tube is fitted into an outer vessel made of gun metal. This vessel has a wide tube at the top which is fitted with a frictionless piston and a pressure gauge as shown in figure. The solution under study is taken in the outer gun metal vessel. As a result of osmosis, the level in the tube tends to fall. The pressure applied on the solution by means of the piston which keeps the level in the tube at is a measure of the osmotic pressure and can be read directly on the pressure gauge.

This method is superior to the other methods because of the following reasons :

(a) In this method, the osmotic pressure is balanced by the external pressure so that there is no strain on the membrane.

(b) The concentration of the solution does not change because the entry of the solvent into the solution is prevented by the external pressure.

(c) The time taken for the measurement of osmotic pressure is much less in this method as compared to the other methods.

Q. Draw a suitable labelled diagram to express the relationships for ideal solution of A and B between vapour pressure and mole fractions of components at constant temperature.

[CBSE 2002; Delhi 2002]

Sol.

Relationships between V.P. and mole fraction for ideal solution

Q. $2 g$ of benzoic acid $\left(C_{6} H_{5} C O O H\right)$ dissolved in $25 g$ of benzene shows a depression in freezing point equal to $1.62 K .$ Molal depression constant for benzene is $4.9 K$ $k g \mathrm{mol}^{-1} .$ What is the percentage association of acid if it forms dimer in solution?

Sol. The given quantities are : $w_{2}=2 g ; K_{f}=4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$ ;

$w_{1}=25 g, \Delta T_{f}=1.62 K$

Substituting these values in equation

$M_{2}=\frac{k_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$

$M_{2}=\frac{4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} \times 2 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{25 \mathrm{g} \times 1.62 \mathrm{K}}=241.98 \mathrm{gmol}^{-1}$

Thus, experimental molar mass of benzoic acid in benzene is $=241.98 \mathrm{g} \mathrm{mol}^{-1}$

Now consider the following equilibrium for the acid:

$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$ BHA $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)_{2}$

If $x$ represents the degree of association of the solute then we would have $(1-x)$ mol of benzoic acid left in

unassociated form and correspondingly $\frac{x}{2}$ as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is:

$1-x+\frac{x}{2}=1-\frac{x}{2}$

Thus, total number of moles of particles at equilibrium equals Van’t Hoff factor $i$

But $i=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}=\frac{122 g \mathrm{mol}^{-1}}{241.98 \mathrm{gmol}^{-1}}$

$\Rightarrow \frac{x}{2}=1-\frac{122}{241.98}=1-0.504=0.496$

$\Rightarrow x=2 \times 0.496=0.992$

Therefore, degree of association of benzoic acid in benzene is $99.2 \%$

Solid State | Question Bank for Class 12 Chemistry

Get Solid State important questions for Boards exams. Download or View the Important Question bank for Class 12 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 12 chemistry chapter wise CBSE.

Click Here for Detailed Chapter-wise Notes of Chemistry for Class 12th, JEE & NEET.

You can access free study material for all three subject’s Physics, Chemistry and Mathematics.

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Also get to know about the strategies to Crack Exam in limited time period.

Q. What is the C.N of $\mathrm{Ca}^{2+}$ and $\mathrm{F}^{-}$ in $\mathrm{CaF}_{2}$ crystal lattice?

Sol. C.N of $C a^{2+}$ ion $=8,$ C.N. of $F^{-}$ ion $=4$

Q. How many particles are present in a face centred unit cell?

Sol. Four particles are present in each unit cell.

Q. How many atoms are there in a unit cell of a metal crystallising in fcc structure ?

Sol. Four

Q. A compound $\mathrm{AB}_{2}$ possesses the $\mathrm{CaF}_{2}$ type crystal structure. Write the coordination numbers of $\mathrm{A}^{2+}$ and $\mathrm{B}^{-}$ ions in its crystals.

Sol. Coordination no. of $\mathrm{A}=8,$ Coordination no. of $\mathrm{B}=4$

Q. What other element may be added to silicon to make electrons available for conduction of an electric current ?

Sol. Phosphorus.

Q. Define coordination number of a metal ion in an ionic crystal.

Sol. In an ionic crystal, coordination number of a metal ion (+ve ion) is the number of negative ions surrounding the metal ion, i.e., which are present as its nearest neighbours.

Q. At what temperature ferrimagnetic substance changes to paramagnetic ?

Sol. 850 K

Q. What is the co-ordination number of $\mathrm{Na}^{+}$ and $\mathrm{Cl}$ – ions in $\mathrm{NaCl}$ structure?

Sol. Six.

Q. Sodium chloride has a face centred cubic crystal. What is the co-ordination number of sodium in sodium chloride ?

Sol. Six

Q. What are the co-ordination numbers of each of the ions present in the cubic close packed structure of $\mathrm{Na}_{2} \mathrm{O}$ at ordinary temperature and pressure?

Sol. $N a=4, O=8$

Q. A metallic element crystallises into a lattice containing a sequence of layers of ABABAB $\ldots . .$ Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space?

Sol. 26%

Q. AgI crystallises in cubic close packed ZnS structure. What fraction of tetrahedral sites are occupied by Ag” ions?

Sol. In the face-centred unit cell, there are eight tetrahedral voids. Of these, half are occupied by silver cations.

Q. The unit cell of a substance has cations $\mathrm{A}^{+}$ at the corners of the unit cell and the anions $\mathrm{B}^{-}$ in the centre. What is the simplest formula of the substance?

Sol. AB

Q. What is the maximum possible co-ordination number of an atom in an hcp crystal structure of an element?

Sol. 12

Q. In a crystal of zinc sulphide, zinc occupies tetrahedral voids. What is the co-ordination number of zinc ?

Sol. 4.

Q. Which out of $\mathrm{Cd} \mathrm{Cl}_{2}$ and $\mathrm{NaCl}$ will produce Schottky defect if added to $\mathrm{AgCl}$ crystal $?$

Sol. $\mathrm{CdCl}_{2}$ will produce Schottky defect.

Q. What type of stoichiometric defects are noticed in crystals ?

Sol. These are of two types i.e., Frenkel and Schottky defects.

Q. What type of semi-conductors is produced when silicon is doped with arsenic ?

Sol. It is of n-type because one electron of each silicon atom remains non-bonded.

Q. What is the effect of Frenkel defect on the electrical conductivity of crystalline solids ?

Sol. It increases because of the vacancies that are created.

Q. Why does Frenkel defect not change the density of AgCl crystals?

Sol. Due to Frenkel defect, as no ions are missing from the crystal as a whole, so there is no change in density.

Q. Mention one property which is caused due to the presence of Fcentre in a solid.

Sol. F-centre is reponsible for the colour and paramagnetic behaviour of the solid.

Q. What type of crystal defect is produced when sodium chloride is doped with MgCl_?

Sol. It is called impurity defect. A cation vacancy is produced. A substitutional solid solution is formed (because $2 \mathrm{Na}^{+}$ ion are replaced by one Mg $^{2+}$ ion in the lattice site).

Q. What causes the conduction of electricity by semiconductors ?

Sol. Electrons and holes produced by defects.

Q. A cubic solid is made up of two atoms A and B. Atoms A are present at the corners and B at the centre of the body. What is the formula of the unit cell ?

Sol. Contribution by the atoms A present at eight corners = 1

Contribution by the atom B present in the centre of the body = 1

Thus, the ratio of atoms of A : B = 1 : 1

Formula of unit cell = AB

Q. Define the term ‘amorphous’. Give a few examples of amorphous solids.

Sol. An amorphous solid is a substance whose constituent particles do not possess a regular orderly arrangement. Some examples of amorphous solids are glass, plastics, rubber, starch etc.

Q. Calculate the density of silver, which is known to crystallise in face-centred cubic form. distance nearest metal atoms is $287 \mathrm{pm}$ (Molar mass of $\left.\mathrm{Ag}=107.87 \mathrm{g} \mathrm{mol}^{-1} ; \mathrm{N}_{0}=6.022 \times 10^{23} \mathrm{mol}^{-1}\right)$

Sol. Density of unit cell $(\rho)=\frac{Z \times M}{N_{0} \times a^{3} \times 10^{-30}}$

According to available data,

Distance between nearest metal atom $(2 r)=287 \mathrm{pm}$

Edge length for foc crytal $(a)=\sqrt{2} \times 2 r=405.87 \mathrm{pm}$

No. of atoms per unit cell $(z)=4$

Atomic mass of silver $(M)=107.87 \mathrm{gmol}^{-1}$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

Density of silver

$=\frac{4 \times 107.87 g m o l^{-1}}{(405.87)^{3} \times 6.022 \times 10^{23} \mathrm{mol}^{-1} \times 10^{-30} \mathrm{m}^{3}}$

$=10.77 \mathrm{g} \mathrm{cm}^{-3}$

Q. What is the distance between $N a^{+}$ and $C I^{-}$ ions in $N a C l$ crystal if its density is $2.165 \mathrm{g} \mathrm{cm}^{-3} ?$ crystallises in fec lattice?

Sol. Step I. Calculation of Edge Length of unit cell

Let the edge length of unit cell $=a$

Volume of unit cell $=a^{3}$

Gram formula mass of $N a C l=23+35.5=585 g m o l^{1}$

No. of particles in foc type unit cell $(Z)=4$

Mass of unit cell $=\frac{Z \times \text { Gramformulamassof } N a C l}{\text { Avogadro’s Number }\left(N_{0}\right)}$

$=\frac{4 \times\left(58.5 g \mathrm{mol}^{-1}\right)}{\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right)}=3.886 \times 10^{-22} \mathrm{g}$

Density of unit cell $(\rho)=2.165 g \mathrm{cm}^{-3}$

Now, Density of unit cell $=\frac{\left(3.886 \times 10^{-22} g\right)}{\text { Volume of unit cell }\left(a^{3}\right)}$

volume of unit cell $(a)=\frac{\left(3.886 \times 10^{-22} g\right)}{\left(2.165 g \mathrm{cm}^{-3}\right)}$

$=1.795 \times 10^{-22} \mathrm{cm}^{3}=179.5 \times 10^{-24} \mathrm{cm}^{3}$

Edge length $(\mathrm{a})=\left(179.5 \times 10^{-24} \mathrm{cm}^{3}\right)^{1 / 3}=5.64 \times 10^{-8} \mathrm{cm}$

$=564 \times 10^{-10} \mathrm{cm}=564 \mathrm{pm}$

Step II. Calculation of distance between $N a^{+}$ and $C l^{-}$ ions.

Edge length of $N a^{+} C l^{-}$ unit cell $=2\left(r_{N a^{+}}+r_{C l}^{-}\right)$

$2\left(r_{N a^{+}}+r_{C I^{-}}\right)=564 \mathrm{pm}$

or $\left(r_{N a^{+}}+r_{C l^{-}}\right)=\frac{564}{2}=282 \mathrm{pm}$

$\therefore$ Distance in and ions $=282$ pm

Q. Calculate the value of Avogadro’s number from the following data : Density of $\mathrm{NaCl}=2.165 \mathrm{g} \mathrm{cm}^{-3},$ Distance between $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ ions in $\mathrm{NaCl}$ crystal $=281 \mathrm{pm}$

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{30}}$ or $\quad N_{0}=\frac{Z \times M}{a^{3} \times \rho \times 10^{-30}}$

Edge length of

$N a C l(a)=2\left(r_{N a^{+}}+r_{C l}\right)=2 . \times 281=562 \mathrm{pm}$

No. of atoms per unit cell $(Z)=4$

$(\because N a C l$ hasfcc structure)

Density of $N a C l(\rho)=2.165 g \mathrm{cm}^{-3}$

Gram molecular mass of $N a C l(M)=23+35.5$

$=58.5 g \mathrm{mol}^{-1}$

$\therefore$ Avogadro’s Number

$\left(N_{0}\right)=\frac{4 \times\left(58.5 g m o l^{-1}\right)}{(562)^{3} \times\left(2.165 g \mathrm{cm}^{-3}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=6.089 \times 10^{23} \mathrm{mol}^{-1}$

Q. How does the electrical conductivity of metallic conductors vary with temperature ?

Sol. Electrical conductivity decreases with increase in temperature because kernels start vibrating and create hinderance in the flow of electrons.

Q. What is the coordination number of

(i) Sodium in sodium oxide $\left(\mathrm{Na}_{2} \mathrm{O}\right)$

(ii) Oxide ion in sodium oxide (Na,O)?

(iii) Calcium in calcium fluoride $\left(\mathrm{CaF}_{2}\right) ?$

(iv) $\operatorname{zinc}$ in zinc blende (ZnS)?

Sol. (i) 4 (ii) 8 (iii) 8 (iv) 4 (v) 4.

Q. In CaF- $_{2}$ crystal $\mathrm{Ca}^{2+}$ ions are present in FCC arrangement. Calculate the number of $\mathrm{F}^{-}$ ions in the unit cell.

Sol. No. of $C a^{2+}$ ions per unit cell $=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4 .$ Hence,

no. of $F^{-}$ ions per unit cell $=2 \times 4=8$

Q. How many atoms can be assigned to its unit cell if an element forms (i) body-centred cubic cell (ii) a face-centred cubic cell.

Sol. (i) 2 (ii) 4.

Q. What are the types of lattice imperfections found in crystals ?

Sol. (a) Stoichiometric defects, viz., Schottky defect and Frenkel defect and (b) Non-stoichiometric defects, viz., metal excess, metal deficiency and impurity defects.

Q. What are non-soichiometric compounds ?

Sol. If the actual ratio of the cations and anions is not same as represented by the ideal chemical formula of the compound, it is called a non-stoichiometric compound.

Q. What is Frenkel defect ?

Sol. When some ions (usually cations) are missing from the lattice sites and they occupy the interstitial sites so that electrical neutrality as well as stoichiometry is maintained, it is called Frenkel defect

Q. Explain with the help of diagrams the structural differ-ences between three types of cubic crystals.

Sol. The crystals belonging to cubic system have three kinds of Bravais lattices. These are :

(i) Simple cubic lattice : There are points only at the corners of each unit.

(ii) Face-centred cubic lattice : There are points at the corners as well as at the centre of each of the six faces of the cube.

(iii) Body-centred cubic lattice : There are points at the corners as well as in the body centre of each cube.

Q. What is the difference between ‘Schottky defects’ and ‘Frenkel detects’ ?

Sol.

Q. Pure silicon is an insulator. Silicon doped with phosphorus is a semiconductor. Silicon doped with gallium is also a semiconductor. What is the difference between the two doped silicon semiconductors ?

Sol. When silicon is doped with phosphorus (a group 15 element) one electron of each phosphorus atom becomes free because it is not involved in any bonding with silicon atom. Thus, silicon containing phosphorus as the impurity is n-type semi-conductor. When silicon is doped with gallium (a group 13 element) only three valence electrons of silicon may be involved in the bonding with the three valence electrons of gallium. Since silicon atom has four valence electrons, this may result in creating vacancies or holes. These holes responsible for electrical conductivity and will result in p-type semi-conductors.

Q. Explain superconductivity.

Sol. There are certain solids called super conducting solids which do not offer any resistance to flow of electric current which means that they have zero resistivity. A few important examples are

$B a_{7} K_{3} B i O_{3}, L a_{1.8} S r_{.2} C u O_{4}, Y B a_{2} C u_{3} O_{7}$ etc.

Q. Sodium crystallises in a body-centred cubic unit cell (bcc) with edge length 4.29 $\hat{\mathrm{A}}$. What is the radius of the sodium atom? What is the length of the body-diagonal of the unit cell?

Sol. For a body centred cubic unit cell (bcc)

Q. Aluminium metal forms a cubic face centred close packed crystal structure. Its atomic radius is

$125 \times 10^{-12} \mathrm{m}$

(a) Calculate the length of the side of the unit cell.

(b) How many unit cells are there in $1.0 \mathrm{m}^{3}$ of aluminium?

Sol. (a) For a face centred cubic lattice (fcc).

Radius $(\mathrm{r})=\frac{a}{2 \sqrt{2}}$

$a=r \times 2 \sqrt{2}=125 \times 10^{-12} \times 2 \sqrt{2} m$

$=125 \times 2 \times 1.414 \times 10^{-12}=354 \times 10^{-12} \mathrm{m}$

(b) Volume of unit cell $(a)^{3}$

$=\left(354 \times 10^{-12}\right)^{3} m^{3}=4.436 \times 10^{-29} m^{3}$

No. of unit cells in $1.0 \mathrm{m}^{3}$ of $\mathrm{Al}$

$=\frac{\left(1.0 m^{3}\right)}{\left(4.436 \times 10^{-29} m^{3}\right)}=2.25 \times 10^{28}$

Q. The edge length of $\mathrm{NaCl}$ unit cell is 564 pm. What is the density of $\mathrm{NaCl}$ in $\mathrm{g} / \mathrm{cm}^{3} ?$

Sol. Density of unit cell $(\rho)=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

According to available data,

Edge length $(a)=564$ pm

Molar mass of $N a C l(M)=23+35.5=58.5 \mathrm{g} \mathrm{mol}^{-1}$

No. of atoms per unit cell” $(Z)=4$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

Density of unit cell

$=\frac{4 \times\left(58.5 g m o l^{-1}\right)}{(564)^{3} \times\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=2.16 g \mathrm{cm}^{-3}$

Q. Cesium chloride crystallises as a body centred cubic lattice and has a density of $4.0 \mathrm{gcm}^{-3} .$ Calculate the edge length of unit cell of cesium chloride crystal.

(Moalr mass of $\left.\mathrm{CsCl}=168.5 \mathrm{g} \mathrm{mol}^{-1} ; \mathrm{N}_{\mathrm{A}}=6.02 \times 1023 \mathrm{mol}^{-1}\right)$

Sol. Density of unit cell $(\rho)=\frac{Z \times M}{N_{0} \times a^{3} \times 10^{-30}}$

or $\quad a^{3}=\frac{Z \times M}{\rho \times N_{0} \times 10^{-30}}$

According to available data,

No. of atoms in the b.c.c. unit cell $(Z)=2$

Density of unit cell $(\rho)=4.0 \mathrm{g} \mathrm{cm}^{-3}$

Molar mass of $\operatorname{CsCl}(M)=168.5 \mathrm{g} \mathrm{mol}^{-1}$

Avogadro’s no. $\left(N_{0}\right)=6.02 \times 10^{23} \mathrm{mol}^{-1}$

By substituting the values,

$a^{3}=\frac{2 \times\left(168.5 g m o l^{-1}\right)}{\left(4.0 g c m^{-3}\right) \times\left(6.02 \times 10^{23} \mathrm{mol}^{-1}\right) \times 10^{-30}}$

$=1.399 \times 10^{8}(p m)^{3}$

Edge length $(a)=\left[1.399 \times 10^{8}(p m)^{3}\right]^{1 / 3}$

(.: Value of a is in pm)

$=1.12 \times 10^{2} \mathrm{pm}$

Q. How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell ? Explain

Sol. (a) The number of nearest neighbours in a packing is called coordination number. For example, in hexagonal close packing (HCP) in three dimensions each atom (or constituent particle) is touching 12 other atoms. So its coordination number is 12 .

(b) $\quad$ (i) In cubic close-packed structures each atom is in direct contact with 12 other atoms. Hence, its coordination number is 12

(ii) In body centred cubic structure coordination number is $8 .$

Q. ‘Stability of a crystal is reflected in the magnitude of its melting point.’ Comment. Collect melting point of solid water, ethyl alcohol, diethyl ether, and methane from a data book. What can you say about the intermolecular forces between these molecules.

Sol. A solid having stronger interparticle forces has higher melting point and is more stable. Thus, we can get an idea about stability of a crystal from its melting point. More stable crystals have higher melting points than less stable crystal. Thus, it is correct to say that stability of a crystal is reflected in magnitude of its melting point.

Methane < Diethyl ether < Ethyl alcohol < Water

Water having highest melting point has the strongest interparticle forces whereas methane having lowest melting point has the weakest intermolecular forces.

The intermolecular forces in these molecules are in the order :

Methane < Diethyl ether < Ethyl alcohol < Water

Q. How will you distinguish between the following pairs of terms

(a) Hexagonal close packing and cubic close packing

(b) Crystal lattice and unit cell

(c) Tetrahedral void and octahedral void.

Sol. (a) In hexagonal close packing every third layer is similar to the first layer. Thus there is ABAB ………..arrangement. In cubic close packing every fourth layer is similar to first thus, there is ABC ABC ….. arrangement.

(b) A crystal lattice may be defined as a regular three dimensional arrangement of identical points in space. While a unit cell may be defined as a three dimensional group of lattice points that generates the whole lattice by translation or stacking.

(c) The vacant space between four touching sphere is called tetrahedral void. Centres of these spheres occupy corners of regular tetrahedron.

The interstitial void formed by combination of two triangular voids of the first and second layer is called octahedral void because this is enclosed between six spheres, centres of which occupy corners of a regular octahedron.

Q. What is a semiconductor ? Describe the two main types of semiconductors and contrast their conduction mechanisms.

Sol. Semi-conductors. These are the solids whose conductivity lies in-between those of conductors and insulators. It normally ranges between $10^{-6}$ to $10^{4}$ ohm $^{-1} \mathrm{cm}^{-1} .$ They, infact allow only a portion of applied electric current to pass through them.

Conduction of electricity in semi-conductors

The electric conductivity of semi conductors is explained with the help of band theory. As we know, there is only a small energy gap between valence band (filled band) and conduction band (empty band) in semi-conductors. Therefore, some electrons can jump from valence band to vacant band and electric conductance is thus, possible. It can further increases with the rise in temperature since more electrons can jump to the conduction band.

Intrinsic semi-conductors. The extrinsic semi-conductors are formed when impurities of certain elements are added to the insulators.

This process known as doping makes available electrons or holes for conductivity and thus, leads to two types of semi-conduction. These are called n-type semi-conductors and p-type semi-conductors.

n-type semi-conductors. n-type semiconductors are formed when impurity atoms containing more valence electrons than the atoms of the parent in insulator are introduced in it. These are called electron rich impurities. For example, when traces of phosphorus (a group 15 element) is added to pure silicon (a group 14 element) on electron of each phosphorus atom (has five valence electrons) becomes free because it is not involved in any bonding with silicon atom (has four valence electrons). The unbonded electrons are free to carry the electric current. Thus, silicon containing phosphorus as the impurity is n-type semi-conductor.

(ii) p-type semi-conductors. These are formed when impurity atoms containing lesser number of valence electrons than the atoms of the parent insulator element added to it. These are called electron deficient impurities. For example, when traces of boron. (a group 13 element) are added to pure silicon (a group 14 element) only three valence electrons of silicon may be involved in the bonding with the three valence electrons of boron. Since silicon atom has four valence electrons, this may result in creating vacancies or holes. These holes are responsible for electrical conductivity and will result in p-type semi-conductors.

Q. Classify each of the following as being either a p-type or an n-type semiconductor :

(i) Ge doped with In

(ii) B doped with Si

Sol. (i) Ge is an element of group- 14 and has configuration $4 s^{2} 4 p^{2}$ It has been doped with In, a $13^{\text {th }}$ group element having $5 s^{2} 5 p^{1}$ configuration. All three valence electrons of In get bonded with three out of four electrons of Ge. Thus, the fourth bond of Ge contains only one electron and hence is an electron deficient bond or a hole. Conductivity is due to these holes or electron deficient bonds. Therefore, it is a p-type semiconductor.

(ii) Boron, $\mathrm{B}$ is an element of group- 13 and has $2 s^{2} 2 p^{1}$ configuration. It is doped with $S i,$ an element of group-

14 having $3 s^{2} 3 p^{2}$ configuration. All three electrons of boron get bonded with 3 out of 4 electrons of Si and $4^{\text {th }}$ electron of impurity atom (i.e., Si) is responsible for conductivity. Thus it is a n-type semiconductor.

Q. An element crystallises in a bcc structure. The edge length of its unit cell is 288 pm. If the density of the crystal is $7.3 \mathrm{g} \mathrm{cm}^{-3}$, what is the atomic mass of the element?

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

or $M=\frac{\rho \times a^{3} \times N_{0} \times 10^{-30}}{Z}$

According to available data,

Edge length $(a)=288$ pm

No. of atoms per unit cell $(Z)=2$

$(\because \text { bcc structure })$

Density of the element $=7.3 \mathrm{g} \mathrm{cm}^{-3}$

Avogadro’s No. $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

$=52.51 g \mathrm{mol}^{-1}$

Q. The element chromium crystallises in a body centred cubic lattice whose density is $7.20 \mathrm{g} / \mathrm{cm}^{3} .$ The length of the edge of the unit cell is $288.4 \mathrm{pm} .$ Calculate Avogadro’s number (Atomic mass of $\mathrm{Cr}=52)$

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

Or

$N_{0}=\frac{Z \times M}{a^{3} \times \rho \times 10^{-30}}$

Edge length of the unit cell $(\mathrm{a})=288.4 \mathrm{pm}$

No. of atoms per unit cell $(Z)=2$

Density of the unit cell $(\rho)=7.2 \mathrm{g} / \mathrm{cm}^{3}$

Atomic mass of the element $(\mathrm{M})=52 \mathrm{g} \mathrm{mol}^{-1}$

$N_{0}=\frac{2 \times\left(52 g m o l^{-1}\right)}{(288.4)^{3} \times\left(7.2 g \mathrm{cm}^{-3}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=6.04 \times 10^{23} \mathrm{mol}^{-1}$

Q. The density of chromium is $7.2 \mathrm{g} \mathrm{cm}^{-3} .$ If the unit cell is a cubic with edge length of 289 pm, determine the type of the unit cell. (Atomic mass of $\mathrm{Cr}=52$ amu)

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

or $\quad Z=\frac{\rho \times a^{3} \times N_{0} \times 10^{-30}}{M}$

Gram atomic mass of $C r(M)=52.0 \mathrm{gmol}^{-1}$

Edge length of unit cell $(\mathrm{a})=289 \mathrm{pm}$

Density of unit cell $(\rho)=7.2 \mathrm{g} \mathrm{cm}^{-3}$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

$\therefore Z=\frac{\left(7.2 g \mathrm{cm}^{-3}\right) \times\left(289^{3} \times\left(6.022 \times 10^{23} \mathrm{mol}^{1}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)\right.}{\left(520 \mathrm{gmol}^{1}\right)}=2$

since the unit cell has 2 atoms, it is body centre in nature.

Q. Calculate the efficiency of packing in case of a metal crystal for

(a) Simple cubic (b) Body centred cubic

(c) Face centred cubic

Sol. (a) Packing Efficiency of Simple Cubic Structure

In a simple cubic unit cell there is only one atom per unit cell.

Let a be the edge length of the unit cell and r be the radius of sphere.

Volume of the sphere $=\frac{4}{3} \pi r^{3}$

As the spheres at the corners are touching each other, the edge length a is equal to $2 \mathrm{r}$.

Volume of the cube $=a^{3}=(2 r)^{3}=8 r^{3}$

$\%$ of the space occupied by spheres

$=\frac{\text { Volume of sphere }}{\text { Volume of cube }} \times 100=\frac{\frac{4}{3} \pi r^{3}}{8 r^{3}} \times 100$

Thus, packing efficiency of simple cubic lattice is 52.4%

(b) Packing Efficiency of Body Centred Cubic (BCC) Structure

We know that in a body centred cubic unit cell there are two spheres per unit cell. Let a be the side of the unit cell and r be the radius of sphere.

volume of sphere $=\frac{4}{3} \pi r^{3}$

Volume occupied by two spheres $=2 \times \frac{4}{3} \pi r^{3}=\frac{8}{3} \pi r^{3}$

Face diagonal $A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{2 a^{2}}$

In right angled triangle ACD

Body diagonal $A D=\sqrt{A C^{2}+C D^{2}}=\sqrt{2 a^{2}+a^{2}}=\sqrt{3} a$

In a body centred cubic unit cell, the spheres at the corners are not touching each other but are in contact with the sphere at the centre. As a result the body diagonal AD is equal to the four times the radius of the sphere.

$\therefore A D=\sqrt{3} a=4 r$

$a=\frac{4}{\sqrt{3}} r$

Volume of unit cell $=a^{3}=\left(\frac{4}{\sqrt{3}} r\right)^{3}=\frac{64}{3 \sqrt{3}} r^{3}$

Percentage of space occupied by spheres

$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100$

$=\frac{\frac{8}{3} \pi r^{3}}{\frac{64}{3 \sqrt{3}} r^{3}} \times 100=\frac{8}{3} \times \frac{22}{7} \times \frac{3 \sqrt{3}}{64} \times 100=68 \%$

Thus, packing efficiency of a bcc arrangement is $68 \% .$

(c) Face centred cubic

In a cubic close packing, the unit cell is face centred cube.

In a face centred cubic unit cell there are 4 spheres per unit cell.

Let r be the radius of sphere and a be the edge length of the cube.

Volume of sphere $=\frac{4}{3} \pi r^{3}$

Volume of four spheres $=4 \times \frac{4}{3} \pi r^{3}=\frac{16}{3} \pi r^{3}$

In a face centred cubic unit cell, the spheres at corners are in contact with sphere at the face centre but are not touching each other. Therefore, face diagonal $\mathrm{AC}$ is equal to four times the radius of sphere.

$A C=4 r$

But in right angled triangle ABC (figure)

$A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$

$\therefore \sqrt{2} a=4 r$

$a=\frac{4 r}{\sqrt{2}}$

Volume of the unit cell $=a^{3}=\left(\frac{4 r}{\sqrt{3}}\right)^{3}=\frac{64}{2 \sqrt{2}} r^{3}$

Percentage of space occupied by spheres

$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100=\frac{\frac{16}{3} \pi^{3}}{\frac{64}{2 \sqrt{2}} r^{3}} \times 100$

$=\frac{16}{3} \times \frac{22}{7} \times \frac{2 \sqrt{2}}{64} \times 100=74 \%$

Thus, packing efficiency of ccp arrangement is 74%. Similarly, packing efficiency of hcp is also 74%.

Q. Explain the following terms with suitable examples :

(i) Schottky defect

(ii) Frenkel defect

(iii) Interstitials

(iv) F-centres

Sol. (i) Schottky Defects. This type of defect is created when one positive ion and one negative ion are missing from their respective positions leaving behind a pair of holes (figure.). Schottky defect are more common in ionic compounds with high co-ordination number, and where the sizes of positive and negative ions are almost equal. For example, $N a C l, K C l, C s C l, A g B r$ and $K B r .$

$\mathrm{NaCl}$ has one defect in $10^{15}$ lattice sites at room temperature. The number of defects increases with increases in temperature.

The number of defects increases to one in $10^{6}$ sites at $775 \mathrm{K}$

and one in $10^{4}$ sites at $1075 \mathrm{K} .$ The presence of large number of Schottky defects in crystal results in significant decrease in its density. This defect is also known as vacancy defect.

(ii) Frenkel Defect. This type of defect is created when an ion leaves its correct lattice site and occupies an interstitial site (fig. Frenkel defects are common in ionic compound which have low co-ordination number and in which there is large difference in size between positive and negative ions. For example,

$Z n S, A g C l, A g B r$ and $A g I$

(iii) Interstitial Defects. This type of defect is caused due to the presence of ions in the normally vacant interstitial sites in the crystal. The ions occupying the interstitial sites are called interstitials. The formation of interstitial defects in determined by the size of the interstitial ion.

(iv) The “holes’ occupied by electrons are called F-centres (or colour centres) and are responsible for the colour of the compound and many other interesting properties. For example, the excess sodium in $\mathrm{NaCl}$ makes the crystal appear yellow, excess potassium in $K C l$ makes it violet and excess lithium in LiClmakes it pink. Greater the number of F-centres, greater is the intensity of colour. Solids containing F-centres are paramagnetic because the electrons occupying the “holes’ are unpaired.

Q. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is $125 \mathrm{pm} .$

(a) What is the length of the side of the unit cell?

(b) How many unit cells are there in $1.00 \mathrm{cm}^{3}$ of aluminium?

Sol. (a) In cubic close packed structure, face diagonal of the unit cell is equal to four times the atomic radius

Face diagonal $=4 \times r=4 \times 125 \mathrm{pm}=500 \mathrm{pm}$

But face diagonal $=\sqrt{2} \times$ edge length

$\therefore$ Edge length $=\frac{\text { Face diagonal }}{\sqrt{2}}=\frac{500}{\sqrt{2}}=354 \mathrm{pm}$

(b) Volume of one unit cell

$=\mathrm{a}^{3}$

$=\left(3.54 \times 10^{-8} \mathrm{cm}\right)^{3}$

No. of unit cells $1.00 \mathrm{cm}^{3}$

$=\frac{1.00}{\left(3.54 \times 10^{-8}\right)^{3}}$

$=2.26 \times 10^{22}$

Chemical Kinetics Solved Problems Class 12 Chemistry

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Aldehydes Ketones and Carboxylic Acids Class 12 Important Questions & Answers

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Q. Give the IUPAC names of:

Sol. (a) Diacetone alcohol, (b) Crotonaldehy de.

(a) $\quad 4$ – Hy droxy – 4 -methylpentan-2-one

(b) $\quad$ But- 2 -en- 1 -al.

Q. Name one reagent used to distinguish acetaldehyde from acetone.

Sol. Tollen’s reagent or Fehling’s solution.

Q. Name one reagent used to convert toluene into benzaldehyde.

Sol. Chromyl chloride in $\mathrm{CS}_{2}\left(\mathrm{CrO}_{2} \mathrm{Cl}_{2} / \mathrm{CS}_{2}\right)$ or chromic trioxide in acetic anhydride $\left[\mathrm{CrO}_{3} /\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\right]$ followed by acid or alkaline hydrolysis.

Q. What type of aldehydes and ketones undergo aldol condensation?

Sol. Aldehydes and ketones containing $\alpha$ -hydrogens.

Q. Which type of aldehydes undergo Cannizzaro reaction?

Sol. Aromatic and aliphatic aldehydes which do not contain $\alpha$ hydrogens

Q. Which alkene on reductive ozonolysis gives acetone as the only product?

Sol.

Q. Write the IUPAC name of

Sol. 2-Methylpentan-3-one

Q. To what oxidation state does ethanal reduce $C u(\mathrm{II}) ?$

Sol. +1 oxidation state

Q. Write the structure and give the IUPAC names of the following compounds. (a) Cinnamic acid (b) Crotonic acid (c) Oxalic acid.

Sol.

Q. Why does benzoic acid not undergo Friedel-Crafts reaction ?

Sol. (a) Due to deactivation of the benzene ring by electronwithdrawing effect of the $-C O O H$ group.

$(b)$ $A I C l_{3}$ gets bonded to the $-C O O H$ group.

Q. Why carboxylic acid do not form oximes ?

Sol. Due to resonance between lone pairs of electrons on the -atom of the group and the carboxyl carbon is less electrophilic than carbonyl carbon in aldedhydes and ketones. Therefore, nucleophilic addition of to the group of carboxylic acids does not occur and hence carboxylic acids do not form oximes.

Q. Give a suitable example of Hell-Volhard Zelinsky reaction.

Sol.

Q. Carbonyl compounds are more polar than alcohols although electronegativity difference between C and O atoms is less than O and H atoms. Explain.

Sol. In the carbonyl group $(>C=O),$ the $\pi$ -electron pair is loosely held and can be readily shifted to the oxygen atom. It is not the case with the alcoholic $(O-H)$ group. Therefore, carbonyl compounds are more polar and have higher dipole moment values $(2.3-2.8 D)$ than the alcohols $(1.6-1.8 D)$

Q. Give the IUPAC name of (a) Isobutyraldehy de (b) Acrolein (c) Valeraldehyde.

Sol.

Q. Which alkene upon reductive ozonolysis will give acetone as the product ?

Sol. 2, 3-dimethylbut-2-ene will form acetone by reductive ozonolysis

Q. Formaldehyde gives cannizzaro’s reaction while acetaldehyde does not. Why ?

Sol. Formaldehyde does not have an $\alpha$ -hydrogen. It therefore, takes part in cannizzaro’s reaction when heated with strong $N a O H$ to form a methyl alcohol and sodium formate mixture. Acetaldehyde fails to react since it has $\alpha$ -hydrogens present.

Q. What happens when ethylbenzene is heated with acidified $K_{2} C r_{2} O_{7} ?$

Sol. Benzoic acid is formed.

Q. Why is benzoic acid a stronger acid than acetic acid?

Sol. The $K_{a}$ value of benzoic acid $\left(6.3 \times 10^{-5}\right)$ is more than that of

acetic acid $\left(1.75 \times 10^{-5}\right) .$ Actually, $C_{6} H_{5}$ group with $-I$

effect in facilitates the release of $H^{+}$ from benzoic acid while

$\mathrm{CH}_{3}$ group with $+I$ effect tends to retard it.

Q. Which out of acetic and peracetic acid is a stronger acid and why ?

Sol. Acetic acid $\left(p K_{a}=4.8\right)$ is stronger than peracetic acid

$\left(p K_{a}=8.2\right) .$ It is because after losing $H^{+}$ ion acetic acid molecule gets stabilised by resonance while the peracetate ion does not.

Q. Carboxylic acids do not give the characteristic reactions of carbonyl group.

Sol. Due to the presence of lone pairs of electrons on the oxygen atom of the OH group, the carboxylic acids are stabilized by resonance.

As a result, the double bond character of the bond in carboxylic acids is greatly reduced as compared to that in aldehydes and ketones. In other words, the carbonyl group in carboxylic acids is not a true carbonyl group as in aldehydes and ketones and hence does not give some of the characteristic reactions of the carbonyl group. For example, unlike aldehydes and ketones, carboxylic acids do not form oximes, hydrazones, semicarbazones etc.

Q. Write reactions and conditions for following conversions :

(a) Phenol to salicylic acid

(b) $\quad 2$ -propanone into 2 -methyl- 2 -propanol.

Sol. (a) By Kolbe’s reaction. Sodium salt of phenol reacts with

$\mathrm{CO}_{2}$ under high pressure of $4-7$ atm, at about $400 \mathrm{K}$ to form salicylic acid.

Q. Write chemical tests to distinguish between formic acid and acetic acid.

Sol.

Q. Although both $>\mathrm{C}=\mathrm{C}<$ and $>\mathrm{C}=\mathrm{O}$ have a double bond, they exhibit different type of addition reactions. Explain.

Sol. The $>\mathrm{C}=\mathrm{C}<$ undergoes electrophilic addition reactions while $>\mathrm{C}=\mathrm{O}$ shows nucleophilic addition reactions. This difference in behaviour is due to the different shapes of their $\pi$ -electron clouds. The electron cloud of bond due to similar electronegativities of the two carbon atoms is almost symmetrical and surrounds both the carbon atoms equally. Consequently, if a reagent is to attack one of the carbon atoms of the bond, it has to pass through the electron cloud. since the -electron cloud consists of loosely held electrons, therefore, it can readily allow electrophiles to react. As a result, the typical reactions of are electrophilc addition reactions. In contrast, the electron cloud of the $\succ \mathrm{C}=0$ is unsymmetrical i.e., shifts towards oxygen due to greater electronegativity of O than C. As a result, the C-atom of the bond acquires a partial charge and hence is readily attacked by nucleophiles. Thus, the typical reactions of are nucleophilic addition reactions.

Q. What is the function of Roschelle salt in Fehling’s solution?

Sol. In alkaline medium, $C u^{2+}$ ions get precipitated as $C u(O H)_{2}$ To keep ions in solution in alkaline medium, Rochelle salt is added. The insoluble first formed goes into solution due to the formation of a soluble complex between ions and tartrate ion as shown below:

Q. Suggest a suitable oxidising agent for the conversion $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCOCH}_{3} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCO}_{2} \mathrm{H}$

Sol. Alkaline $K M n O_{4},$ acidified $K_{2} C r_{2} O_{7}$ or $H N O_{3}$ cannot be used since all of these will cleave the molecule at the site of double bond giving a mixture of ketone/acids. The most suitable reagent for this oxidation is $\mathrm{NaOI} \quad\left(I_{2} / N a O H\right)$ since methyl ketones on treatment with NaOI undergo iodoform reaction to give iodoform along with the salt of a carboxylic acid having one carbon atom less than the starting methyl ketone.

Q. Fluorine is more electronegative than chlorine even than $p-$ fluorobenzoic acid a weaker acid than $p$ -chlorobenzoic acid. Explain

Sol. since halogens are more electronegative than carbon and also possess lone pairs of electrons, therefore, they exert both $-I$ and $+R$ -effects. Now in $F,$ the lone pairs of electrons are present in $2 p$ -orbitals but in $C l,$ they are present in $3 p-$ orbitals.

since $2 p$ -orbitals of $F$ and $C$ are of almost equal size, therefore, the $+R$ -effect is more pronounced in $p-$ fluorobenzoic acid than in -chlorobenzoic acid

Thus, in $p$ -fluorobenzoic acid, $+R$ -effect outweighs the $-I-$ effect but is p-chlorobenzoic acid, it is the I-effect which outweighs the $+$ R-effect. Consequently acid is a weaker acid than p-chlorobenzoic acid.

Q. Give an example of a reaction where a Grignard reagent acts as a reducing agent.

Sol. If the Grignard reagent or the ketone contains branching at the $\alpha$ -carbon, sometimes the usual addition reaction does not occur due to steric hindrance. Instead reduction occur in which a hydride ion is transferred from the Grignard reagent to the ketone via a six-membered cyclic transition state in a manner similar to

M.P.V. reduction. For example, when isopropylmagnesium bromide is added to di-isopropyl ketone, the expected $3^{\circ}$ alcohol (i.e., tri-isopropylcarbinol) is not formed; instead the secondary alcohol, di-isopropylcarbinol is obtained by reduction of the keto group.

Q. Boiling points of carbonyl compounds lie between the parent alkanes and corresponding alcohols. Justify.

Sol. Alkanes are non-polar in nature and the attractive forces are weak vander Waal’s forces. In alcohols, intermolecular hydrogen bonding is present in the molecules. In the carbonyl compounds, only dipolar forces exist in the molecules. It is, therefore, quite obvious that the attractive forces in carbonyl compounds are more than in alkanes and less in comparison to alcohols. This justifies the trend in boiling point values in the members of these families.

Q. Halogen acids readily combine with alkenes to form addition products but fail to react with carbonyl compounds. Discuss.

Sol. Halogen acids (HX) react with alkenes to form haloalkanes but they fail to react with the carbonyl compounds. In fact, the attack does take place but the product is unstable and decomposes to form carbonyl compound along with halogen acid. Thus, the reaction is reversible.

Q. Out of benzaldehyde and propionaldehye which is more reactive towards nucleophilic addition ?

Sol. Propionaldehyde is more reactive than benzaldehyde towards nucleophilic addition. Actually, in benzaldehyde the aldehydic group has (mesomeric) effect. As a result, the electron density on the carbonyl carbon atom tends to increase and the nucleophile attack is more difficult as compared to propanal where alkyl group is attached to the aldehydic group.

Q. Chloral hydrate is a gem-diol but still stable. How will you account for it?

Sol. In chloral hydrate, the presence of three $C l$ atoms with $-I$ effect increases the magnitude of the positive charge on the carbonyl carbon atom. As a result, even a weak nucleophile like $H_{2} O$ attacks to form a hydrate. Moreover, there is weak intermolecular hydrogen bonding in the chloral hydrate which tends to increase its stability.

Q. Why does not formaldehyde take part in aldol condensation?

Sol. Aldol condensation involves the nucleophile attack of carbanion generated by one molecule of a particular carbonyl compound on the other molecule. For this, the carbonyl compound must have atleast one $\alpha$ -hydrogen present. since formaldehyde $(H C H O)$ has no such hydrogen present, it fails to take part in aldol condensation alongwith another carbonyl compound with atleast one -hydrogen atom e.g., formaldehyde and acetaldehyde.

Q. Why is it necessary to control the pH during the reaction of aldehydes and ketones with ammonia derivatives?

Sol. The addition of ammonia derivative $\left(N H_{2}-G\right)$ to aldehydes and ketones is done in weakly acidic medium (pH about 3.5). In case the medium is strongly acidic (pH close to 1 ), then the ammonia derivative will be also protonated and will not be able to act as a nucleophile.

Q. Benzophenone does not react with $\mathrm{NaHSO}_{3} .$ Explain.

Sol. The addition of $N a H S O_{3}$ on benzophenone $\left[\left(C_{6} H_{5}\right)_{2} C=O\right]$ involves the nucleophile attack of bisulphite $\left(H S O_{3}^{-}\right)$ ion. The phenyl groups act as hindrance to the attacking nucleophile. Therefore, benzophenone does not react with $\mathrm{NaHSO}_{3}$.

Q. Arrange the following in decreasing ease of hydration:

(a) $\mathrm{CH}_{3} \mathrm{CHO},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}, \mathrm{HCHO}, \mathrm{CH}_{2} \mathrm{ClCOCH}_{3}, \mathrm{CH}_{2} \mathrm{ClCHO}$

Sol. Aldehydes are more easily hydrated than ketones. Further, the presence of electron withdrawing groups increases the ease of hydration of a carbonyl compound. In the light of this, the correct order is :

(a) $\mathrm{HCHO}>\mathrm{CH}_{2} \mathrm{ClCHO}>\mathrm{CH}_{3} \mathrm{CHO}>\mathrm{CH}_{2} \mathrm{ClCOCH}_{3}>\mathrm{CH}_{3} \mathrm{COCH}_{3}$

Q. Write the mechanism for the following conversion.

Sol.

Phenylacetylene

Q. Carboxylic acids do not give the characteristic reactions of carbonyl group. Explain.

Sol. The carbonyl group in carboxylic acid is not a free group as in aldehydes and ketones. It is involved in resonance.

Therefore, carboxylic acids fail to give the characteristic reactions of the carbonyl groups as given by aldehydes and ketones.

Q. Formic acid reduces Tollen’s reagent while other carboxylic acids do not. Justify.

Sol. $\mathrm{HCOOH}$ has an aldehydic (CHO) group in addition to

carboxyl group $(\mathrm{COOH}) .$ Therefore, it is expected to behave as reducing agent also and reduces Tollens to form a shining mirror.

Q. 2, 4, 6-trimethylbenzoic acid is quite difficult to esterify. Assign reason.

Sol.

In the trimethylbenzoic acid there are three electron releasing $\mathrm{CH}_{3}$ groups present. They tend to increase the electron density on the oxygen atom of $O-H$ bond in carboxylic acid. As a result, the acidic strength is reduced considerably and it is not so easy to esterify $2,4,6$ -trimethyl benzoic acid.

Q. m-Hydroxybenzoic acid is a stronger acid than benzoic acid while p-hydroxybenzoic acid is weaker. Explain.

Sol. The hydroxyl $(O H)$ group has both $-I$ effect and $+R$ effect. Whereas the former helps in the release of $H^{+}$ from the carboxylic group, the latter tends to oppose the same because of resonance. At the para position, the $+\mathrm{R}$ effect dominates the effect. Therefore, p-hydroxy benzoic acid is a weaker acid than benzoic acid. But in case of m-hydroxybenzoic acid, the $+\mathrm{R}$ effect does not operate to the same extent as in para isomer. Thus, the acidic weakening effect is smaller and meta hydroxy benzoic acid is a stronger acid than benzoic acid.

Q. Carbon-oxygen bond lengths in formic acid are different but are the same in sodium formate. Justify.

Sol. In sodium formate, the contributing structures for the anion are equivalent while these are not the same in formic acid.

Thus, carbon-oxygen bond length in formate ion is the same for both the bonds while these are different in formic acid.

Q. Phenate ion has more number of contributing structures than benzoate ion; but still benzoic acid is a stronger acid. Explain.

Sol. In phenate ion, the negative charge available is dispersed only one electrongative oxygen atom while there are two oxygen atoms in benzoate ion to disperse the negative charge.

This means that benzoate ion is more stable than phenate ion and this makes benzoic acid stronger acid than phenol although the contributing structures for phenate ion (five) are more than for benzoate ion (two).

Q. Tertiary butyl benzene does not give benzoic acid when oxidised with $K M n O_{4} .$ Why?

Sol. In order that an alkyl group attached to benzene ring may be oxidised to carboxyl group, the presence of atleast one

$\alpha$ -hydrogen is necessary. For example, toluene is oxidised to benzoic acid. But tertiary butyl benzene has no -hydrogen atom. It is therefore, not oxidised to benzoic acid.

Q. $\mathrm{CH}_{3} \mathrm{COO}^{-}$ ion is more stable than $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$ ion. Assign reason.

Sol. In acetate ion, the negative charge gets dispersed due to the electron withdrawing nature of carbonyl group. On the other

hand, in ethoxide ion, the $C_{2} H_{5}$ group with $+I$ effect tends to increase the electron density on the ion and destabilise it.

Q. Which out of each pair is expected to be a stronger acid ?

(a) $\quad \mathrm{CH}_{3} \mathrm{COOH}$ or $\mathrm{HCOOH}$

(b) $\quad \mathrm{CH}_{3} \mathrm{COOH}$ or $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$

(c) $\quad C_{6} H_{5} \mathrm{COOH}$ or $\mathrm{HCOOH}$

(d) $\quad C H_{2}(C l) C O O H$ or $C H_{2}(B r) C O O H$

Sol. (a) $\mathrm{HCOOH}$

(b) $\mathrm{CH}_{3} \mathrm{COOH}$

(c) HCOOH

(d) $\mathrm{CH}_{2}(\mathrm{Cl}) \mathrm{COOH}$

Q. How will you prepare 2-methylbutanoic acid from butan-2-ol ?

Sol.

Q. Complete following :

Sol.

Q. How will you synthesise the following by using a Grignard reagent?

(a) $2,2$ -Dimethylpentanoic acid (b) But-3-enoic acid

Sol.

Q. Why is acetyl chloride a better acetylating agent than acetic acid ?

Sol. The acetylation is carried by acetyl carbocation $\left(\mathrm{CH}_{3} \mathrm{CO}^{+}\right)$ The cation is formed by acetyl chloride and not by acetic acid.

Therefore, acetyl chloride is a better acetylating agent than acetic acids.

Q. Draw the structural formula of hex-2-en-4-ynoic acid

Describe the following giving a chemical equation for each

(i) Transesterification,

(ii) Hofmann bromamide reaction

Sol. (i) Transesterification : This reaction involves the nucleophilic acyl substitution of the alkoxy group of an ester with the alkoxy group of an alcohol in the presence of an acid or alkali

$R C O O R^{\prime}+R^{\prime \prime} O H \longrightarrow R C O O R^{\prime \prime}+R^{\prime} O H$

(ii) Hofmann bromamide reaction: This reaction involves the conversion of an amide into primary amine with one carbon atom less than the parent amide, by reaction with bromine and alkali

Q. Predict the products of the following reactions :

Sol.

(c) $\quad \mathrm{NH}_{2}$ attached to $\mathrm{NH}$ is more nucleophile than $\mathrm{NH}_{2}$ attached to $\mathrm{C}=$ O group, therefore, reaction occurs through $\mathrm{NH}_{2}$ attached to NH to give the corresponding semicarbozone, i.e.,

Q. Which acid of each pair shown here would you expect to be stronger ?

Sol.

Q. Although phenoxide ion has more number of resonating structure than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Sol. In the resonating structure of phenoxide ion the negative charge is dispersed on only one oxygen atom whereas in the carboxylate ion the charge gets dispersed on two oxygen atoms. Thus, the dispersal of charge is more in the case of carboxylate ion and hence the resonance stabilisation is more. This is why the carboxylic acids are more stronger than phenol (s).

Q. Give reasons for the following :

(i) Aldehydes are more reactive than ketones in nucleophilic reactions.

(ii) Most aromatic acids are solids but the acids of acetic acid group are mostly liquids.

Sol. (i) The reactivities of aldehydes is more than that of ketones due to the following reasons:

(a) Inductive effect: The ease with which a nucleophile attacks the carbonyl group depends upon the electron deficiency i.e., the magnitude of the positive charge on carbonyl carbon. since an alkyl group has electron donating inductive effect ( $+$ I effect), therefore greater the number of alkyl groups attached to carbonyl carbon greater is the electron density on the carbonyl carbon and hence lower is its reactivity. Aldehydes are more polar than ketones.

In ketone we have two alkyl groups whereas in aldehydes only one alkyl group is attached to carbonyl carbon. Since there is lesser number of alkyl groups in case of aldehydes so they undergo nucleophillic addition more readily than ketones.

(b) Steric effect : The presence of more alkyl groups in case of ketones hinders the attack of nucleophile on the carbonyl group and so the ketones are less reactive than aldehydes.

(ii) Aromatic acids contain benzene ring and their molecular masses are much higher than aliphatic acids of acetic acid group. Aromatic, acids also involve stronger inter molecular attractions, thus they are solids.

Q. Write IUPAC names of following compounds :

Sol.

$2-(2-$ bromophenyl) ethanal. It can also be named as

$2-(\text { o-bromophenyl })$ ethanal.

(b) 5 -chloro-3-ethyl-2-pentanone.

(c) $3,5$ -dimethyl phenyl ethanoate

Q. Give reasons for the following :

(a) Acetic acid is a weaker acid than chloroacetic acid.

(b) During preparation of ammonia derivatives of aldehydes and ketones, $p H$ of medium is controlled.

Sol. (a) The acidic nature of a carboxylic acid is due to its ability to release the $H^{+}$ (proton). Any factor that stabilise the carboxylate ion would facilitate the release of proton and thus increase the acid strength. The electron withdrawing group (attached to $-\mathrm{COOH}$ gp) increases the acid strength. Due to electron withdrawing effect $(-I \text { effect })$ of chlorine atom, the negative charge of carboxylate ion formed gets dispersed. As such this ion gets stabilised and is, therefore, more easily formed.

The acetate ion obtained from acetic acid gets detabilised due to electron releasing effect of methyl group and is not formed

easily. Consequently dissociation of $\mathrm{CH}_{3} \mathrm{COOH}$ takes place to lesser extent.

(b) Formation of ammonia derivatives (oximes, hydrazone, semi-carbazone etc.) proceeds via the attack of carbonyl carbon with proton to form the conjugate acid.

Therefore, presence of an acid is a must for preparing these

derivatives $(p H<7)$ However, in strongly acid medium, the proton attacks the unshared pair of electrons on nitrogen to form the species

$R N^{+} H_{3}$ which cannot attack the carbonyl carbon.

$\stackrel{+}{H}+: N H_{2} R \longrightarrow N H_{3} R$

In basic medium, there is no protonation of carbonyl group and

hence no reaction.

$>C=O \frac{\text { basic medium }}{\text { medium }} \rightarrow$ No protonation Therefore, preparation of ammonia derivatives requires slightly

acidic medium $(p H=3.5)$ and its careful control is essential.

Q. How does $N H_{3}$ react with

(a) Formaldehyde,

(b) Acetaldehyde

(c) Semicarbazide

Sol. (a) $N H_{3}$ reacts with formaldehyde to form hexamethylene tetramine (also called urotropine).

Q. (a) What is ammonolysis of esters?

(b) How will you convert:

(i) Acetaldehyde to acetamide

(ii) Methanol to acetic acid?

Sol. (a) Esters react with ammonia to form amides. The reaction is called ammonolysis of esters.

Q. Addition of Grignard reagents to dry ice followed by hydrolysis gives carboxylic acids while that of organolithium compounds under similar conditions gives ketones. Explain

Sol. since electronegativity of $L i(\mathrm{E} . \mathrm{N} .=1.0)$ is lower than that of $M g(\mathrm{E.N.}=1.2),$ therefore, organolithium compounds are more nucleophilite than Grignard reagents. As a result, organolithium compounds not only add to the more reactive $\mathrm{CO}_{2}$ but also to the less reactive resonance stabilized lithium salt of carboxylic acid thus formed to yield ketones.

Grignard reagents, on the other hand, being less nucleophilic

add only to the more reactive $\mathrm{CO}_{2}$ but not to the less reactive resonance stabilized magnesium salt of the carboxylic acid from which carbxylic acid can be generated by hydrolysis with mineral acids.

Q. How will you prepare methanol from formaldehyde without using a reducing agent.

Sol. Cannizzaro reaction is a disproportionation reaction in which one molecule of an aldehyde is reduced while the other is oxidised.

Here a conc. solution of $N a O H$ is used which is not a reducing agent.

Q. You are provided with following four reagents. $I_{2} / N a O H$ $\mathrm{NaHSO}_{3},$ LiAlH_{4} Schiff’s reagent. Write which two reagents can be used to distinguish between the compounds in each of the following pairs:

(i) $\quad \mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{CH}_{3} \mathrm{COCH}_{3}$

(ii) $\quad \mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$

(iii) $\quad C_{6} H_{5} C O C H_{3}$ and $C_{6} H_{5} C O C_{6} H_{5}$

Sol. (i) $\quad \mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ can be distinguished by :

LiAIH$_{4}$ and Schiff’s reagent.

(ii) $\mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$ can be distinguished by $\mathrm{I}_{2} / \mathrm{NaOH}$ and Schiff’s reagent (Benzaldehy de restores pink colour to Schiff’s reagent very slowly).

(iii) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COC}_{6} \mathrm{H}_{5}$ can be distinguished by

$\mathrm{NaHSO}_{3}$ and $\mathrm{I}_{2} / \mathrm{NaOH}$

Q. An organic compound [A] molecular formula $C_{4} H_{8} O$ when reduced with $\mathrm{NaBH}_{4}$ gives compound [\textrm{B} ] \text { which reacts with } HBr to form compound [C][\mathrm{ optically active } ] \text { Identify } \mathrm { A } , \mathrm { B } , \mathrm { C }

and write the two enantiomers of compound $C$.

Sol. The data suggests that the given compound [A] is an aliphatic

ketone butanone $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\right) .$ The reactions involved are listed.

Q. Two moles of an ester $[\mathrm{A}]$ are condensed in the presence of sodium ethoxide to give a $\beta$ -ketoester $[\mathrm{B}]$ and ethanol. On heating in an acidic solution $[\mathrm{B}]$ gives ethanol and a $\beta$ -ketoacid $[\mathrm{C}] .$ On decarboxylation, [C] gives pentan- 3 -one. Identify $[\mathrm{A}],[\mathrm{B}]$ and $[\mathrm{C}]$ with proper reasoning.

Sol. (a) According to available data [C] is a $\beta$ -keto acid which upon decarboxylation gives pentane- 3 -one. The structure of is :

$[\mathrm{C}]$

(b) The compound $[\mathrm{C}]$ along with ethanol has been obtained from compound [B] upon acid hydrolysis. since $[\mathrm{B}]$ is a $\beta$ ketoester, its structure may be given as follows:

(c) since $[\mathrm{B}]$ has been formed by the condensation of two moles of $[\mathrm{A}]$ in the presence of sodium ethoxide, the ester $[\mathrm{A}]$ is ethyl propanoate and it participates in Claisen Condensation as follows:

Q. Give chemical tests to distinguish between the following pairs of compounds :

(a) Propanal and Propanone

(b) Methyl acetate and ethyl acetate

(c) Benzaldehyde and Benzoic acid

Sol. (a) When propanal is added to tollen’s reagent (an ammoniacal solution of silver nitrate), silver oxide is reduced to metallic silver which deposits as a mirre

(b) When ethyl acetate is treated with aqueous $N a O H$ Sodium ethanoate and ethanol is formed ethanol on treatment with $I_{2}$ and $N a O H$ forms yellow ppt of iodoform.

When methyl acetate is treated with aqueous NaOH sodium ethanoate and Methanol is formed. Methanol does not respond to iodoform reaction

(c) When benzoic acid is warmed with ethyl alcohol and a little concentrated $\mathrm{H}_{2} \mathrm{SO}_{4},$ a fragrant odour of ethyl benzoate is obtained.

Benzaldehyde does not respond to this test.

Q. Write structural formulae and names of the four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde served as nucleophile and which as electrophile.

Sol.

Q. An organic compound (A) (molecular formula $\left.C_{8} H_{16} O_{2}\right)$ was hydrolysed with dilute sulphuric acid to give a carboxylic acid

(B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-rene. Write equations for the reactions involved.

Sol.

Dehydration of C gives but-1-ene i.e., a product with four carbon atoms, therefore the other product B is also with four carbon atoms and has a terminal COOH group.

Accordingly, A is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$

(butyl butanoate)

$\mathrm{B}$ is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ and

$\mathrm{C}$ is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$

Q. Arrange the following compounds in increasing order of their property as indicated:

(a) Acetaldehyde, Acetone, Di-tert-butyl ketone, methyl tertbutyl ketone (reactivity towards $H C N)$

(b) $C H_{3} C H_{2} C H(B r) \operatorname{COOH}, C H_{3} C H(B r) C H_{2} C O O H$

$\left(\mathrm{CH}_{3}\right) \mathrm{CHCOOH}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (acid strength)

(c) Benzoic acid, 4-Nitrobenzoic acid, 3, 4- Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Sol. (a) The reactivity of aldehydes and ketones toward nudeophillic addition depends upon (i) $+$ I effect (ii) steric hinderance. Hence the order is Di-tert-butyl ketone $<$ methyl-tert-butyl ketone < Acetone $<$ Acetaldehy de.

(b) The acidic strength depends upon (i) nature of $+$ I effect

(ii) nature of atom/group attached (iii) position of substituent on the chain. Hence, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$

$<\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{2} \mathrm{COOH}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{COOH}$

(c) $\quad 4$ -Methoxy benzoic acid $<$ Benzoic acid $<4$ Nitrobenzoic acid $<3,4$ -Dinitrobenzoic acid

Q. Give plausibe explanation for each of the following

(a) Cyclohexanone forms cyanohydrin in good yield but 2 ,

$2,6$ -trimethylcyclohexanone does not.

(b) There are two $-N H_{2}$ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(c) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the

water or the ester should be removed as fast as it is formed.

Sol. (a) The carbonyl group in cyclohexanone is highly

polarised and nuleophilic addition of $H^{\delta+}-C N^{\delta-}$ at carboxyl

group $\left(>C^{\delta+}=O^{\delta-}\right)$ takes place easily. centre of nucleophilic attack by $C N^{-}$

The presence of three methyl group (which are electron repelling) reduces the polarity of $>C=O$ group on one hand and offer a steric hindrance to nucleophilic attack of $C N^{-}$ at group. Therefore, trimethyl cyclohexanone does not give good yield i.e., it gives very poor yield.

(b) The study of semicarbazide (NH3 derivative of urea,

reveal that two $\mathrm{N}-\mathrm{H}$ bonds of $\mathrm{NH}_{2}$ group away from group in semicarbazide are comparatively weaker than two $\mathrm{N}-\mathrm{H}$ bond of the other $-N H_{2}$ group closer to $>C=O$ group. Thus it

produces $H^{+}$ easily for protonation of group of aldehyde or ketone.

The nature of reaction, esterification of carboxylic acid, is reversible as ester formed react with water and produces reactants and thus an equilibrium is set up between reactants and products. The removal of ester or water shifts the equilibrium to right according to Le-Chatelier’s principle and more ester is formed.

Q. Predict the organic product of the following reactions

Sol.

Q. (a) How will you convert

(i) Benzoyl chloride to benzaldehyde

(ii) Propanone to 2 -propanol

(iii) Benzoic acid to $m$ -nitrobenzoic acid

(Write the reaction and state the reaction conditions in each case.)

(b) Write the names and structures of the products formed in the following reactions

(i) Reaction of semicarbazide $\left(N H_{2} C O N H N H_{2}\right)$ with formaldehyde.

(ii) Oxidation of ethyl benzene with alkaline $K M n O_{4}$.

Sol. (a) (i) Benzoyl chloride is subjected to Rosenmund reduction.

Q. Account for the following

(i) Chloroacetic acid has lower $p K a$ value than acetic acid

(ii) Electrophillic substitution in benzoic acid takes place at meta position.

(iii) Carboxylic acid have higher boiling points than alcohols of comparable molecular masses.

Sol. (i) Chloroacetic acid is a stronger acid than acetic acid and has higher value of dissociation constant $K a$ than that of acetic acid. We know that $p K a=-\log K a,$ it means that chloroacetic acid having higher value $K a$ will have lower value of $p K a .$ Chloroacetic acid is stronger than acetic acid because $C l$ group is electron withdrawing and chloroacetate ion is more stabilised than acetate ion (as methyl group is electron releasing).

(ii) Carboxyl group (-COOH) is electron withdrawing i.e. deactivating the benzene ring and thus electron density becomes very less at ortho and para position in comparison to meta position. Electrophiles (+vely charged species) find it easier to attack at meta position as there is higher electron density thus $-C O O H$ group is meta directing.

As there is positive charge on ortho and para position, electron density is higher at meta position and hence electrophilic substitution takes place at meta position. (Similarly nitro group is meta directing)

(iii) Carboxylic group (-COOH) in acids is highly polar and carboxylic acid, generally, exists as dimers containing two hydrogen bonds each as shown below.

These hydrogen bonds in carboxylic acids are stronger than those in alcohols. It is due to following two factors

(a) The $O-H$ bond of the carboxylic acids are more strongly polarised due to adjacent electron attracting $>C=$ O groups.

(b) The oxygen atom of the group $C=O$ in carboxylic acid is more negative as compared to the oxygen atom of the alcohol.

Thus carboxylic acids posses higher boiling points than corresponding alcohols of similar molecular masses.

Q. Write chemical reactions for preparation of acetone from the following

(i) $\quad$ A secondary alcohol

(ii) Carboxylic acid

(iii) Acid chloride

(iv) Alkyne

(v) Alkene

Sol. (i) Acetone can be prepared by oxidation or catalytic dehydrogenation of isopropyl alcohol.

Q. How will you convert.

(a) Ethanal to lactic acid

(b) Acetaldehyde to acetone

(c) Ethanal to 2 -hydroxybut- 3 -enoic acid

(d) Formaldehyde to acetaldehyde

(e) Acetaldehyde to formaldehyde

(f) Formaldehyde to $n$ -butane

(g) Ethanol to butan- 2 -one

(h) Acetaldehyde to crotonic acid

(i) Propanal to propyne

(j) Acetone to phorone

(k) Propanone to iodoform

(1) Formaldehyde to urotropine

(m) Acetone to tertiary butyl alcohol

(n) Ethanal to propan- 2 -ol

(o) Methyl cyanide to ethanal

(p) Benzalchloride into cinnamaldehyde.

(q) Propene to propanone

(r) Acetophenone to 2 -phenylbutan- 2 -ol

Sol. (a) Ethanal to lactic acid

Q. Write the equations for the reaction of :

reagents:

(i) Heating with excess of $H B r$

(ii) Dilute $K M n O_{4}$ in the presence of alkali

(iii) A dilute solution of $N a_{2} C O_{3}$

(iv) Phenyl hydrazine

(v) Tollen’s reagent

Sol.

Q. An organic compound (A) with molecular formula $\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}$ forms an orange-red precipitate with $2,4$ -DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollen’s or Fehlings’ reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula $C_{7} H_{6} O_{2} .$ Identify the compounds $(\mathrm{A})$ and $(\mathrm{B})$ and explain the reactions involved.

Sol. (A) forms $2,4$ -DNP derivative. Therefore, it is an aldehyde or a ketone. since it does not reduce Tollen’s or Fehling reagent,

(A) must be a ketone. (A) responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyer’s reagent. This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of $(\mathrm{A})$ indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows:

Q. What is meant by the following terms? Give an example in each case.

(a) Cyanohydrin

(b) Semicarbazone

(c) Hemiacetal

(d) Ketal

(e) $2,4$ -DNP-derivative

(f) Acetal

(g) Aldol

(h) Oxime

(i) Imine

(j) Schiff’s base

Sol. (a) Cyanohydrin is a compound which contains both OH and CN group. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

(b) The product of carbonyl compounds with semicarbazide is known as semicarbazone.

(c) Hemiacetal is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

(d) Ketal is a cyclic compound obtained by reaction of acetone with ethylene glycol.

(e) $2,4$ -DNP derivative: see properties of aldehydes/ ketones in section 12.7 of text.

(f) Acetal : Acetal are the product of aldehyde and monohydric alcohol

(g) Aldol is a condensation product of acetaldehyde in the presence of dil. $N a O H$

(h) Oxime : Aldehyde and Ketones react with hydroxylamine to form oximes.

(i) Schiff’s base : Schiff’s reagent is an aqueous solution of magenta or pink coloured rosaniline hydrochloride which has been decolourised by passing $\mathrm{SO}_{2}$ when aldehyde are treated with decolourised solution of schiff’s reagent, its pink or magenta colour is restored. This reaction is used as a test for aldehyde because ketones do not restore the pink colour of schiff reagent.

Aliphatic Hydrocarbon | Question Bank for Class 12 Chemistry

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Q. What are the natural sources of hydrocarbons ?

Sol. Coal and petroleum are two natural sources of hydrocarbons.

Q. What is petroleum ?

Sol. Petroleum is a complex mixture of solid, liquid and gaseous hydrocarbons.

Q. What is the main constituent of LPG ?

Sol. LPG mainly contains butane and Isobutane.

Q. Give a scale for measuring the quality of gasoline.

Sol. Octane number.

Q. Arrange the following in order of increasing volatility :

gasoline, kerosene oil and diesel.

Sol. Diesel < kerosene oil < gasoline.

Q. Name the compound that is added to LPG to detect its leakage.

Sol. The compound added to LPG to detect its leakage is ethyl

mercaptan $\left(C_{2} H_{5} S H\right)$

Q. Name the product obtained by the fractional distillation of petroleum which is used as a furnace fuel in metallurgical operations.

Sol. Fuel oil is used as furnace fuel in metallurgical operations.

Q. Define knocking.

Sol. A sharp metallic rattling sound produced in an internal combustion engine, is called knocking.

Q. What is octane number of fuel which contains $95 \%$ iso-octaneand $5 \%$ n-heptane?

Sol. Octane number is 95.

Q. What is meant by anti-knocking compound ? Give its example.

Sol. – Those compounds which reduces knocking and help in smooth combustion of gasoline are called antiknocking compounds, e.g. tetra ethyl lead $\left(C_{2} H_{5}\right)_{4} P b$

Q. Why is dipole moment of trans-1, 2 dichloroethene zero ?

Sol. It is because individual dipoles are equal and opposite such that net dipole moment is zero

Q. Why alkanes and cycloalkanes are called paraffins ?

Sol. They are called paraffins because under normal conditions alkanes are inert towards common reagents [In Latin, Parum = little, affinis =affinity]

Q. Why eclipsed and staggered forms of ethane cannot be isolated at room temperature ?

Sol. It is because the difference in their energy is less.

Q. What is Lindlar’s catalyst ? Give its use.

Sol. Lindlar’s catalyst is $P d / B a S O_{4} .$ It is used to convert Alkynes to

alkenes with the help of $H_{2}$

Q. Convert ethylene to ethane.

Sol. $\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \frac{\mathrm{Ni}}{573 \mathrm{K}} \mathrm{CH}_{3}-\mathrm{CH}_{3}$

Q. Geometrical isomers are not possible in alkynes why ?

Sol. Due to their straight linear geometry.

Q. Ethyne treated with ozone followed by treatment with zinc and water, leads to formation of compound ‘X’. What is X ?

Sol. X is glyoxal, i.e., ethane 1, 2-dial.

Q. Write chemical equation for the reaction when ethyne is passed through acetic acid in the presence of $\mathrm{Hg}^{+}$.

Sol. In the presence of $H g^{+},$ ethyne adds acetic acid to give vinyl ester.

Q. What are arenes ? Give general formula of monocyclic arenes.

Sol. – Arenes are aromatic hy drocarbons. $C_{n} H_{2 n-6}$ is general formula for monocyclic arenes.

Q. What is trade name of benzene hexachloride ?

Sol. Gammaxene.

Q. What is electrophile in sulphonation ?

Sol. $\mathrm{SO}_{3} \mathrm{H}^{+}$

Q. In presence of conc. $H_{2} S O_{4},$ conc. $H N O_{3}$ acts as acid or base during nitration of benzene?

Sol. It acts as a base.

Q. Write down the products of ozonolysis of $1,2$ dimethylbenzene as cyclohexa- $1,3,5$ -triene.

Sol. With ozone $o$ -xylene forms triozonide that supports the structure of benzene as cyclohexa- $1,3,5$ triene.

Q. How is petroleum refined using technique of fractional distillation ?

Sol. The crude oil is washed with acidic or basic solution depending upon whether the impurities are basic or acidic in nature. The washed oil is then subjected to fractional distillation by heating it to about $723 K$ and allowing the vapours to pass through a fractionating column. The crude oil is separated into various fractions.

Q. Give the full form of CNG and state its composition.

Sol. CNG is compressed natural gas. It mainly consists of methane, $(85$ to $95 \%$ ). Ethane and propane are also present in small quantities. Natural gas is formed in the earth crust by the decomposition of vegetable-matter lying under water by the action of anaerobic bacteria so mostly it is found along with petroleum deposits.

Q. Discuss the stability of conformers of ethane.

Sol. The staggered conformer of ethane is more stable than eclipsed conformer, because in staggered conformer, the H-atoms are far apart. So, the repulsion is lesser than in eclipsed conformation. The difference in the energy content of staggered and eclipsed

conformation is $12.5 \mathrm{kJ} \mathrm{mol}^{-1}$

The order of stability of conformers is in the following order:

Staggered $>$ gauche (skew) $>$ eclipsed.

Q. Classify the following as Z or E isomers :

Sol. (i) E (ii) E (iii) E (iv) Z

Q. What is cracking ? Describe the methods to complete this process.

Sol. Cracking : “The process of breaking down of hydrocarbons of high molecular mass and higher boiling points into smaller hydrocarbons of low molecular mass and lower boiling points, is termed as cracking.

For Example:

Types of cracking : The process of cracking is of two types:

(i) Thermal cracking.

(ii) Catalytic cracking.

(i) Thermal cracking: When cracking is carried at high temperature, it is called thermal cracking.

(ii) Catalytic Cracking: When cracking is carried at low temperature in the presence of a catalyst such as alumina or silica, is called catalytic cracking.

Q. In the alkane $C H_{3}-C H_{2} C\left(C H_{3}\right)_{2} C H_{2} C H\left(C H_{3}\right)_{2},$ identify $1^{\circ}, 2^{\circ}, 3^{\circ}$ and $4^{\circ}$ carbon atoms and give the total number of $H$ atoms bonded to each one of these.

Sol.

Each $1^{\circ}$ Carbon is attached to 3 hydrogen, Each $2^{\circ}$ carbon is attached to 2 hydrogen atoms. Each tertiary ‘C’ atom is bonded to one and quaternary $^{\epsilon} \mathrm{C}^{\prime}$ is not attached to any hydrogen atom.

All five Carbon atoms are attached to total 15 hydrogen atoms.

TwoCarbon atoms are attached to total 4 hydrogen atoms.

OneCarbon atom is attached to total one hydrogen atom

OneCarbon atom is not attached to any hydrogen atom.

Q. Draw Newman projection formulae of the two staggered Gauch conformations of n-butane.

[NCERT]

Sol.

Q. Between the two conformational isomers of cyclohexane, i.e., chair and boat forms, which one is more stable ?

[NCERT]

Sol. Chair form is more stable than boat form.

Q. Addition of to propene yields bromo-2-propane, while in presence of benzoyl peroxide the same reaction yields

1-bromopropane. Explain and give mechanism.

[NCERT]

Sol.

Q. Why propane has only one eclipsed conformation while butane has three ? Explain and give diagrams.

Sol. Propane has only one eclipsed conformer Butane has three eclipsed forms due to changes in dihedral angles.

(i), (ii) and (iii) Newman eclipsed projection of butane.

Q. Complete the reactions

Sol.

Q. Write IUPAC names for the following molecules :

[NCERT]

Sol. (i) 5 -methyl – hept $1,3,6$ -triene.

(ii) 3 methyl pent $-1$ -ene- 4 -yne.

Q. Draw structure of the six isomeric pentenes, $C_{5} H_{10} .$ Specify as $E$ or $Z$ to each geometrical isomer.

[NCERT]

Sol.

Q. Convert ( i ) propene to 1 -bromopropane (ii) 2 -methyl propene to 2 -chloro $-2$ -methyl propane.

Sol.

Q. Why conjugated dienes undergo 1, 4-additions ? Explain.

Sol. Conjugated diene undergo 1, 4 addition due to more resonance stabilization of intermediate carbocation.

Q. Which of the following polymerizes readily and why ?

(i) Acetylene (ii) Ethene (iii) Buta-1, 3-diene.

[NCERT]

Sol. (iii) Buta-1, 3-diene undergo polymerisation reaction readily because it will form resonance stabilized intermediate.

Q. What are the necessary conditions for any compound to show aromaticity ?

[NCERT]

Sol. A compound is said to aromatic if there is delocalization of $\pi-$ electrons and it has planar structure.

It should follow Huckel’s Rule, i.e., in a conjugated, planar, cyclic system if number of delocalized electrons is $(4 n+2) \pi,^{c} n^{\prime}$ being integer, $i . e ., 1,2,3,$ etc. it will be aromatic, otherwise not, e.g.

Benzene, naphthalene, anthracene have $6,10,14$ electrons in conjugated, planar, cyclic system.

Q. Write down the products of ozonolysis of $1,2$ -dimethyl benzene (o-xylene). How does the result support Kekule structure for benzene?

[NCERT]

Sol.

These products prove the structure I and II (Kekule structures of benzene).

Q. Account for the following order of acidity, Acetylen

e > Benzene> Hexane.

[NCERT]

Sol. $-H-C \equiv C-H$ has $s p$ hybridised carbon which is most electronegative due to $50 \%$ s-character therefore it is most acidic.

In Benzene, each carbon is $s p^{2}$ hybridised carbon $(33 \%$ of character), which is less electronegative than hybridised carbon, therefore, it is less acidic.

In hexane, each $^{c} C^{\prime}$ is hybridised carbon $(25 \%$ s-character), which is least electronegative, therefore, it is least acidic.

Q. Why does benzene undergo electrophilic substitution reaction easily and nucleophilic substitution with difficulty ?

[NCERT]

Sol. Benzene has $\pi-$ electrons, therefore, it can undergo electrophilic substitution reactions easily. Nucleophilic substitution are difficult because nucleophile will be repelled by $\pi$ -electrons.

Q. Write down the products and give mechanism of the following reactions :

[NCERT]

Sol.

since electron density is maximum at $o$ and $p$ -positions $S O_{3}$ will attack at $o$ and $p$ -positions.

$\mathrm{NO}_{2}$ will attack at $o$ and $p$ -positions because electron density is maximum at ortho and $p$ -positions.

Q. Arrange the following set of compounds in order of their decreasing reactivity with an electrophile, $E^{+}$

(i) Chlorobenzene, $2,4$ -dinitrochlorobenzene, p-nitrochlorobenzene

(ii) Toluene, $p-H_{3} C-C_{6} H_{4}-C H_{3}, H_{3} C-C_{6} H_{4}-N O_{2}$ $p-O_{2} N-C_{6} H_{4}-N O_{2}$

[NCERT]

Sol.

Q. What happens when-

(i) Sodium acetate is heated with soda lime.

(ii) Acetylene is passed through ammoniacal silver nitrate solution.

(iii) Chlorobenzene is condensed with chloral in the presence of sulphuric acid?

Sol. (i) When sodium acetate is heated with soda lime, methane gas will be formed

(ii) When acetylene is passed through ammonical silver nitrate

solution, white precipitate of silver acetylide will be formed.

(iii) When chlorobenzene is condensed with chloral, in presence of conc. $H_{2} S O_{4},$ DDT is formed.

Q. (i) Write a chemical equation for Friedel-Craft’s reaction.

(ii) Identify $A, B, C$ and $D$ in the following reactions:

Sol. (i) Friedel Crafts Reaction

Q. Bring out following conversions:

(i) 2 -Bromo-propane to 1 Bromo-propane

(ii) Methane to Ethane

OR

An unknown alkene $A$ on reductive ozonolysis gives two isomeric carbonyl compounds $B$ and $C$ of molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$. Write the structure of $A, B$ and $C$

Sol.

Q. What is LPG ? State its composition, properties and applications.

Or

Give the full form of LPG and state its composition. Why it is considered a good fuel ?

Sol. Petroleum gas : It is a mixture of butane, propane, propene, isobutane and ethane. The main constituent of petroleum gas is butane.

At ordinary pressure butane $\left(C_{4} H_{10}\right),$ Propane $\left(C_{3} H_{8}\right)$ and ethane $\left(C_{2} H_{6}\right)$ are gases. Under high pressure, they are easily liquefied. The petroleum gas which has been liquefied under pressure is called liquefied petroleum gas (LPG). The domestic gas cylinder contains LPG.

LPG a better domestic fuel : Components of LPG burn easily and produce a lot of heat, so LPG is used as a good domestic fuel. Due to the following qualities, LPG is considered as a better fuel.

(i) $\mathrm{LPG}$ is an efficient fuel as it has high calorific value, burning of 1 gram of LPG in air produces about 50 kilo joule of heat energy.

(ii) LPG burns with a smokeless flame, so causes no pollution.

(iii) It leaves no ash on burning so it is a clean fuel.

(iv) On burning no poisonous gases are formed.

(v) It is easy to handle and convenient in storing.

Q. Describe the process of fractional distillation of petroleum with the help of a labelled diagram of the distillation tower. Name any four/one products obtained. State the temperature at which each one of them is obtained.

Sol. Petroleum : It is a naturally occurring fossil fuel that is dark coloured, viscous and foul smelling oil. It is a mixture of several hydrocarbons and other elements such as sulphur, oxygen and nitrogen. Mostly it occurs deep down the earth between two nonporous rocks, that’s why it is called as petroleum. (petra =rock, oleum $=$ oil ) or crude – oil.

Principle of refining of petroleum :”The process of separating different components of petroleum, is known as refining. It is carried out by the process of fractional distillation which is based on the fact that different components of petroleum have different boiling point ranges.

The crude-oil is heated in a tank to a temperature of about and all the vapours formed are fed into a tall fractionating column near the bottom. Except asphalt, all the other components are in vapour state. As the mixture of hot vapours rises up in the column, it begins to cool The vapours of components having higher boiling point-range condense first in the lower part and vapours of components with lower boiling point-range condense later at the upper part of column. Therefore, the components with the highest boiling point condenses first and is collected in the lowest part of the column and the component having lowest boiling point condenses last and is collected in the top most part of column. The vapours that do not liquify, are taken out from the top of the column. This component is called petroleum-gas. The residual oil or liquid residue which does not vapourise in the column is called residual oil. It is collected in the base of column and subjected to further distillation.

Products of petroleum : The various components obtained at different heights of fractional column, (in order of top to bottom of column) are as follows :

Q. Give the products when the following reagents react with alkenes (i) Sulphuric acid (ii) halogens (iii) oxidation (iv) ozone.

Sol. (i) Reaction with sulphuric acid: Cold concentrated sulphuric

-acid reacts with alkenes to gives addition products

(ii) Reaction with halogen : Halogens (chlorine and bromine only react at ordinary temperature and without exposure to $U V$ light to form vicinal halide.

(iii) Oxidation

(a) Oxidation with dilute aqueous potassium permanganate solution : At low temperature alkenes form vicinal glycols.

[This reaction is called hydroxylation as hydroxyl groups are attached across the double bond. $]$

(b) With alkaline potassium permanganate at higher temperature : In such conditions cleavage of $C=C$ bond occurs forming carboxylic acids.

(iv) Ozone: With ozone alkenes form ozonides which undergo cleavage to produce carbonyl compounds. This method was extensively used to determine the position of double bond in alkenes in past.

Procedure : Ozone is passed through a solution of an alkene in inert solvent such as $\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}$ or $\mathrm{CCl}_{4}$ at temperature $196-200 K .$ Alkenes are oxidised to ozonised which are not usually isolated but reduced in situ, with zinc dust and water or $H_{2} / P d$ to give carbonyl compound. The location of double bond is assigned on the basis of products.

(a) In case of symmetrical alkene, only one aldehyde is formed on reduction with zinc and water:

$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}+\mathrm{O}_{3} \longrightarrow$

(b) If double bond is terminal, methanal is one product and a higher aldehyde is other product.

$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{O}_{3} \longrightarrow$

(c) If ketone is formed, branching is there at C atom having double bond.

Q. An alkene on ozonolysis gives 2 -butanone and 2 methylpropanal. What products will be obtained when it is treated with hot conc. $K M n O_{4} ?$

Sol. Let us first try to write the structure of alkene. The product of ozonolysis are :

Q. (i) What is Octane number? How does it atfect knocking?

(ii) What type of isomerism is shown by methoxy methane and ethanol?

(iii) Complete the following reactions:

Sol. (i) Octane number is defined as percentage of iso-octane in the mixture of iso-octane and $n$ -heptane which gives same knocking as sample fuel. Higher the octane number, lesser will be knocking.

(ii) Functional isomerism.

Q. Complete the following reactions :

(i) Isopropyl bromide $\frac{\text { alc. KOH }}{\text { Heat }} A \frac{H B r}{\text { Peroxide }} B$

(ii) $\quad n$ -Propyl alcohol $\frac{\text { Conc. } H_{2} S O_{4}}{443 K} A \frac{O_{2}, A g}{\text { Heat }}>B$

(iii) $1,1,2,2,$ -tetrachloroethane $\frac{z n, \text { alcohol }}{\text { Heat }}$

$A \frac{\text { Iron tube }}{675 \mathrm{K}}>B$

(iv) Acetylene $\frac{N a N H_{2}}{\longrightarrow} A \stackrel{C H_{3} C H_{2} B r}{\longrightarrow} B$

(v) Propyne $\frac{H_{2}, P d, B a S O_{4}}{\text { Quinoline }} \rightarrow A \frac{(i) O_{3}}{(i i) Z n, H_{2} O} B$

Sol.

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Q.

Why are substances like platinum and palladium often used for carrying out electrolysis of aqueous solutions?

Sol. Platinum and palladium form inert electrodes, i.e., they are not attacked by the ions of the electrolyte or the products of electrolysis. Hence, they are used as electrodes for carrying out the electrolysis.

Q. Why are powdered substances more effective adsorbent than their crystalline forms ?

Sol. Powdered substances have greater surface area as compared to their crystalline forms. Greater the surface area, greater is the adsorption.

Q. Why is it necessary to remove CO when ammonia is obtained by Haber’s process ?

Sol. CO acts as a poison for the catalyst used in the manufacture of ammonia by Haber’s process. Hence, it is necessary to remove it.

Q. Why is the ester hydrolysis slow in the beginning and becomes faster after some time ?

Sol. The ester hydrolysis takes place as follows :

The acid produced in the reaction acts as catalyst (autocatalyst) for the reaction. Hence, the reaction becomes faster after some time.

Q. What is the role of desorption in the process of catalysis?

Sol. Desorption makes the surface of the solid catalyst free for fresh adsorption of the reactants on the surface.

Q. Give reason why a finely divided substance is more effective as an adsorbent ?

Sol. Adsorption is a surface phenomenon. Since finely divided substance has large surface area, hence, adsorption occurs to a greater extent.

Q. What is demulsification? Name two demulsifiers.

Sol. The decomposition of an emulsion into constituent liquids is called demulsification.

Demulsification can be done by boiling or freezing.

Q. What do you understand by activity and selectivity of catalysts.

Sol. Activity of a catalyst refers to the ability of catalyst to increase the rate of chemical reaction. Selectivity of a catalyst refers to its ability to direct the reaction to give a specific product.

Q. Why does physisorption decrease with increase of temperature ?

Sol. Physisorption is an exothermic process :

According to Le Chatelier’s principle, if we increase the temperature, equilibrium will shift in the backward direction, i.e., gas is released from the adsorbed surface.

Q. What modification can you suggest in the Hardy Schulze law ?

Sol. According to Hardy Schulze law, the coagulating ion has charge opposite to that on the colloidal particles. Hence, the charge on colloidal particles is neutralized and coagulation occurs. The law can be modified to include the following :

When oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, coagulation of both the sols occurs.

Q. Why is it essential to wash the precipitate with water before estimating it quantitatively ?

Sol. Some amount of the electrolytes mixed to form the precipitate remains adsorbed on the surface of the particles of the precipitate. Hence, it is essential to wash the precipitate with water to remove the sticking electrolytes (or any other impurities) before estimating it quantitatively.

Q. Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.

Sol. In adsorption, the substance is concentrated only at surface and does not penetrate through the surface to the bulk of adsorbent while in absorption, the substance is uniformly distributed through out the bulk of the solid.

A distinction can be made between absorption and adsorption by taking an example of water vapour. Water vapours are absorbed by anhydrous calcium chloride but adsorbed by silica gel.

Q. What role does adsorption play in heterogeneous catalysis?

Sol. In heterogeneous catalysis, the reactants are generally gases whereas catalyst is a solid. The reactant molecules are adsorbed on the surface of the solid catalyst by physical adsorption or chemisorption. As a result, the concentration of the reactant molecules on the surface increases and hence the rate of reaction increase. Alternatively, one of the reactant molecules undergo fragmentation on the surface of the solid catalyst producing active species which react faster. The product molecules in either case have no affinity for the solid catalyst and are desorbed making the surface free for fresh adsorption. This theory is called adsorption theory.

Q. Give two chemical methods for the preparation of colloids.

Sol. Colloidal solutions can be prepared by chemical reactions involving, double decomposition, oxidation, reduction and hydrolysis.

(i) Double decomposition : A colloidal sol of arsenious sulphide is obtained by passing hydorgen sulphide into a solution of arsenious oxide in distilled water.

$A s_{2} O_{3}+3 H_{2} S \rightarrow A s_{2} S_{3}+3 H_{2} O$

(ii) Oxidation : A colloidal solution of sulphur can be obtained by passing hydrogen sulphide into a solution of sulphur dioxide in water through a solution of an oxidizing agent (bromide water, nitric acid etc.)

$\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{S} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$

$H_{2} S+[O] \rightarrow H_{2} O+S$

Q. How are the colloidal solutions classified, on the basis of physical states of the dispersed phase and dispersion medium ?

Sol.

Q. Describe a chemical method each for the preparation of sol of sulphur and platinum in water.

Sol. Preparation of sol of sulphur : A colloidal solution o sulphur can be obtained by passing hydrogen sulphide into a solution of sulphur dioxide in water or through solution of an oxidizing agent (bromine water, nitric acid etc.)

$\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{S} \longrightarrow 3 \mathrm{S}+\mathrm{H}_{2} \mathrm{O}$

$H_{2} S+[O] \longrightarrow H_{2} O+[S]$

Preparation of platinum sol : It is prepared by electrical disintegration or Bredig’s arc method. In this method, electric arc is struck between electrodes of the metal immersed in the dispersion medium. The intense heat produced vaporizes some of the metal, which then condenses to form particles of calloidal size.

Q. Give four examples of heterogeneous catalysis.

Sol. When the catalyst exists in a different phase than that of reactants. It is said to be heterogeneous catalyst and the catalysis is called heterogeneous catalysis.

(i) Manufacture of $N H_{3}$ from $N_{2}$ and $H_{2}$ by Haber’s process using iron as catalyst.

$N_{2}+3 H_{2} \stackrel{F e}{\longrightarrow} 2 N H_{3}$

(ii) Manufacture of $\mathrm{CH}_{3} \mathrm{OH}$ from $\mathrm{CO}$ and using $(\mathrm{a}$ mixture of $C u, Z n O$ and $C r_{2} O_{3}$

$\mathrm{CO}+2 \mathrm{H}_{2} \frac{\mathrm{Cu}+\mathrm{ZnO}+\mathrm{Cr}_{2} \mathrm{O}_{3}}{\mathrm{Heat}} \rightarrow \mathrm{CH}_{3} \mathrm{OH}$

(iii) Oxidation of with using as catalyst in ostwald process.

(iv) Hydrogenation of oils to form vegetable ghee using finely divided nickel.

$\mathrm{Oil}+\mathrm{H}_{2} \stackrel{\mathrm{Ni}}{\longrightarrow}$ Vegetable ghee.

Q. Describe some features of catalysis by zeolites

Sol. Features of catalysis by zeolites. (i) Zeolites are hydrated alumino-silicates which have a three-dimensional network structure containing water molecules in their pores. (ii) To use them as catalyst, they are heated so that water of hydration present in the pores is lost and the pores become vacant. (iii) The size of the pores varies from 260 to $740 \mathrm{pm} .$ Thus, only those molecules adsorbed in these pores and catalysed whose size is small enough to enter these pores hence, they act as molecular sieves or shape selective catalyst. An important catalyst used in petroleum industry is ZSM- 5 (Zeolite sieve of molecular porosity 5 ). It converts alcohol into petrol by first dehydrating them to form a mixture of hydrocarbons.

Q. What is shape selective catalysis.

Sol. The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts because of their honeycomb-like structures. They are microporous aluminosilicates with three dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving $A l-O-S i$ framework. The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites. They are found in nature as well as synthesised for catalytic selectivity.

Zeolites are being very widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. An important zeolite catalyst used in the petroleum industry is ZSM- – . It converts alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Q. Give four uses of emulsions.

Sol. Uses of emulsions are :

(1) In medicines : A wide variety of pharmaceutical preparations are emulsions. For example, emulsions of cod liver oil. These emulsified oils are easily acted upon by digestive juices in the stomach and, hence are readily digested.

(2) Digestion of fats : Digestion of fats in the intestines is aided by emulsification.

(3) In disinfectants : The disinfectants such as dettol give emulsions of the oil in water type when mixed with water.

(4) In building roads : An emulsion of asphalt and water is used for building roads. In this way there is no necessity of melting the asphalt.

Q. What are micelles ? Give an example of a micellers system.

Sol. There are some substances which at low concentrations behave as normal strong electrolytes, but at higher concentration exhibit colloidal behaviour due to the formation of aggregates. The aggregated Particles thus formed are called micelles. The formation of micelles takes place only above a particular temperature called kraft temperature $\left(\mathrm{T}_{\mathrm{k}}\right)$ and above a particular concentr-ation called critical micelle concentration (CMC). Surface active agents such as soaps and synthetic detergents belong to this class. For soaps, the CMC is $10^{-4}$ to $10^{-3} \mathrm{moll}^{-1}$. micelles may contain as many as 100 molecules or more.

Q. Comment on the statement that “Colloid is not a substance but state of a substance”.

Sol. The given statement is true. This is because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example, $N a C l$ in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves like a crystalloid while concentrated solution behaves as a colloid (called associated colloid). It is the size of the particles which matters, i.e., the state in which the substance exists. If the size of the particles lies in the range 1 nm to 1000 nm to it is in the colloidal state.

Q. What is the difference between physical adsorption and chemisorption?

Sol.

Q. What are the factors which influence the adsorption of gas on a solid?

Sol. Factors influencing the adsorption of gas on solid:

(i) Nature of gas: Under given conditions of temperature and pressure, the easily liquifiable gases like $N H_{3}, H C l, C O_{2},$ etc. are adsorbed to a greater extent than

the permanent gases such as $H_{2}, O_{2}, N_{2}$ etc. It is because the vander waal’s forces or molecular forces are more predominant in the former than in later category.

Effect of Nature of the Adsorbent: Activated charcoal (ii) is the most common adsorbent for the gases which can be easily liquefied. The gases $H_{2}, O_{2}$ and $N_{2}$ are adsorbed on metals like $N i, P d$ etc.

(iii) Specific area of the solid : Specific area of an adsorbing solid is the surface available for adsorption per gram of the adsorbent. Greater the specific area of the solid, greater would be its adsorbent power.

(vi) Effect of pressure : It is expected that extent of adsorption increases with increase of pressure.

(v) Effect of temperature : In physisorption

$x / m$ decreases with rise in temperature. While in chemisorptionslightly increases in the beginning and then decreases as the temperature rises. This initial increase is due to the fact that, like chemical reactions, chemisorption also requires activation energy.

Sol. Adsorption isotherm. It is a graph indicating the variation of the mass or the gas adsorbed per gram $(x / m)$ of the adsorbent with pressure $(p)$ at constant temperature. Distinction between Freundlich and Langmuir adsorption isotherms

(i) The mathematical expressions representing adsorption are:

Freundlich adsorption isotherm: $\frac{x}{m}=k p^{1 / n}$

Langmuir adsorption isotherm: $\frac{x}{m}=\frac{a p}{1+b p}$

(ii) Langmuir adsorption isotherm is more general and is applicable at all pressure while Freundlich adsorption isotherm fails at high pressure. The latter can be deduced from the former.

Q. What do you understand by activation of adsorbent? How is it achieved?

Sol. Activation of adsorbent implies increases in the adsorption power of the adsorbent. It involves increase in the surface area of the adsorbent and is achieved by following methods.

$\bullet$ By finely dividing the adsorbent.

$\bullet$ By removing the gases already adsorbed

$\bullet$ By making the surface of adsorbent rough by chemical or mechanical methods.

Q. Discuss the effect of pressure and temperature on the adsorption of gases by solids.

Sol. Effect of pressure : It is expected that extent of adsorption increases with increase in pressure. The extent of the adsorption is generally expressed as $x / m$ where $m$ is the mass of the adsorbent and $x$ is that of the adsorbate when equilibrium has been attained. A graph drawn between

extent of adsorption $\left(\frac{x}{m}\right)$ and the pressure $p$ of the gas at constant temperature is called adsorption isotherm.

. Effect of temperature : Magnitude of adsorption decrease with increase in the temperature.

A graph drawn between extent of adsorption $\left(\frac{x}{m}\right)$ and temperature ( $T$ ) at constant pressure is called adsorption isobar physical adsorption isobar shows a decrease in If with rise in temperature, the isobar of chemisorption $x / 1$ shows an increase in the beginning and then decreases as the temperature rises. This initial increase is due to the fact that like chemical reactions, chemisorption also requires activation energy.

Q. What are lyophilic and lyophobic sols ? give one example of each type.

Sol. (i) Lyophilic colloids : The word ‘lyophilic’ means liquid-loving. Colloidal sols directly formed by mixing substances like gum, gelatine, starch, rubber, etc., with a suitable liquid (the dispersion medium) are called lyophilic sols. An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase (say by evaporation), the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. Furthermore, these sols are quite stable and cannot be easily coagulated.

(ii) Lyophobic colloids : The word ‘lyophobic’ means liquid-hating. Substances like metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloids. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are readily precipitated (or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking and hence, are not stable. Further, once precipitated, they do not give back the colloidal sol by simple addition of the dispersion medium. Hence, these sols are also called irreversible sols. Lyophobic sols need stabilising agents for their preservation.

Q. What is the difference between multimolecular and macromolecular colloids ? Give one example of each. How are associated colloids different from these two types of colloids ?

Sol. On dissolution, a large number of atoms or smaller mole-cules of a substance aggregate together to form species having size in the colloidal range (diameter < 1 nm). The species thus formed are multimolecular colloids while macromolecules in suitable solvents form solutions in which the size of macromolecules may be in colloidal range.

Difference between associated colloids multimolecular and macromolecules colloids. Multimolecular colloids are formed by the aggregation of a large number of small atoms molecules such as S8. Macromolecular colloids contain molecules of large size like starch the size of molecules in this case have the dimensions of colloids. Their size lies in the colloidal range. The associated colloids are formed by electrolytes which dissociate into ions and these ions associate together to form ionic micelles whose size lies in the colloidal range, soap is common example of this type . The micelle formation occurs above a particular concentration (called critical micellisation concentration) and above a particular temperature, called Kraft temperature.

Q. Explain what is observed when

(i) An electrolyte is added to ferric hydroxide sol

(ii) An emulsion is subjected to centrifugation

(iii) Direct current is passed through a colloidal sol

(iv) A beam of light is passed through a colloidal solution.

Sol. (i) The positively charged colloidal particles of $F e(O H)_{3}$ sol get coagulated by the oppositely charged ion provided by electrolyte.

(ii) The constituent liquids of the emulsion separate out. In other words, demulsification occurs.

(iii) On passing direct current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

(iv) Scattering of light by the colloidal particles takes place and the path of light becomes illuminated. This is called Tyndall effect.

Q. Explain the following terms :

(1) Peptization, (2) Electrophoresis, (3) Coagulation,

Sol. (1) Peptization : Peptization may be defined as the process of converting a precipitate into colloidal sol by shaking it with dispersion medium in presence of a small amount of electrolyte. The electrolyte used for this purpose is called peptizing agent.

(2) Electrophoresis : In this process colloidal particles move towards oppositely charged electrodes, get discharged and precipitated.

(3) Coagulation : The stability of the lyophobic sol is due to the presence of charge on colloidal particles. If, some how, the charge is removed, the particles will come nearer to each other to form aggregate (or coagulate) and settle down.

Q. Explain the following terms with suitable examples

(1) Gel (2) Aerosol and (3) Hydrosol

Sol. (1) Gel : It is a colloidal dispersion of a liquid in a solid, common example is butter.

(2) Aerosol : It is a colloidal dispersion of a liquid in a gas, common example is, fog.

(3) Hydrosol :It is a colloidal sol of a solid in water as the dispersion medium, common example is starch sol or gold sol.

Q. What are emulsions ? What are their different types ? Give one example of each type.

Sol. These are liquid-liquid colloidal systems, i.e., the dispersion of finely divided droplets in another liquid. If a mixture of two immiscible or partially miscible liquids is shaken, a coarse dispersion of one liquid in the other is obtained which is called emulsion. Generally, one of the two liquids is water. There are two types of emulsions.

(i) Oil dispersed in water (O/W type) and

(ii) Water dispersed in oil (W/O type ).

In the first system, water acts as dispersion medium. Examples of this type of emulsion are milk and vanishing cream. In milk, liquid fat is dispersed in water. In the second system, oil acts as dispersion medium. Common examples of this type are butter and cream.

Emulsions of oil in water are unstable and sometimes they separate into two layers on standing. For stabilisation of an emulsion, a third component called emulsifying agent is usually added. The emulsifying agent forms an interfacial film between suspended particles and the medium. The principal emulsifying agents of O/W emulsions are proteins, gums, natural and synthetic soaps, etc., and for W/O, heavy metal, salts of fatty acids, long chain alcohols, etc.

Emulsions can be diluted with any amount of the dispersion medium. On the other hand, the dispersed liquid when mixed, forms a separate layer. The droplets in emulsions are often negatively charged and can be precipitated by electrolytes. They also show Brownian movement and Tyndall effect. Emulsions can be broken into constituent liquids by heating, freezing, etc.

Q. Action of soap is due to emulsification and micelle formation. Comment.

Sol. Cleansing action of soap: Washing action of soap is due to the emulsification of grease and taking it away in the water along with dirt or dust present on grease.

Explanation : The cleansing action of soap can be explained keeping in mind that a soap molecule contains a non-polar hydrophobic group and a polar hydrophilic group. The dirt is held on the surface of clothes by the oil or grease which is present there. since oil or grease are not soluble in water, therefore, the dirt particles cannot be removed by simply washing the cloth with water. When soap is applied, the non-polar alkyl group dissolves in oil droplets while the polar $-C O O^{-} N a^{+}$ groups remain dissolved in water (Figure). In this way, each oil droplet is surrounded by negative charge. These negatively charged oil droplets cannot coagulate and a stable emulsion is formed. These oil droplets (containing dirt particles) can be washed away with water along with dirt particles.